Applications of Trigonometric Identities – Formulas, Problems, and Step-by-Step Solutions

It is known that $\cos 2A = -\dfrac{7}{25}.$
Because $180^{\circ} \leq 2A \leq 270^{\circ}$, by dividing the three parts by $2$, we obtain $90^{\circ} \leq A \leq 135^{\circ}.$
Therefore, $A$ is in quadrant II.
Notice that $\cos 2A = 2 \cos^2 A-1$ so that we obtain
$\begin{aligned} \cos 2A & = -\dfrac{7}{25} \\ 2 \cos^2 A-1 & = -\dfrac{7}{25} \\ \cos^2 A & = \dfrac{9}{25} \\ \cos A & =- \dfrac35 \end{aligned}$
$\cos A$ is negative because $A$ is in quadrant II.
Known: $\text{adjacent} = 3$ and $\text{hypotenuse} = 5$. Using the right triangle approach and the Pythagorean theorem, we obtain
$\text{opposite} = \sqrt{5^2-3^2} = 4.$
From here, we obtain
$\begin{aligned} \sin A & = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac45 \\ \tan A & = -\dfrac{\text{opposite}}{\text{adjacent}} = -\dfrac43 \\ \csc A & = \dfrac{\text{hypotenuse}}{\text{opposite}} = \dfrac54 \end{aligned}$
Notice that only $\sin$ and its reciprocal, $\csc$, are positive in quadrant II.
Based on the explanation above, the correct answer option is E.

It is known
$\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac35$ and
$\cos B = \dfrac{\text{adjacent}}{\text{hypotenuse}} = – \dfrac35.$
The quadrant where the sine value is positive and the cosine value is negative is quadrant II. Therefore, $A$ and $B$ lie in quadrant II.
Using the right triangle approach and the Pythagorean theorem (Pythagorean Triple: $3, 4, 5),$ we obtain
$\cos A = -\dfrac45$ and $\sin B = \dfrac45.$
Next, by using the angle sum identities:
$\boxed{\begin{aligned} \sin (x + y) & = \sin x \cos y + \cos x \sin y \\ \sin 2x & = 2 \sin x \cos x \\ \cos 2x & = \cos^2 x-\sin^2 x \end{aligned}}$
we obtain
$$\begin{aligned} \sin (2A+B) & = \sin 2A \cos B + \cos 2A \sin B \\ & = (2 \sin A \cos A) \cos B + (\cos^2 A-\sin^2 A) \sin B \\ & = 2 \cdot \dfrac35 \cdot \left(-\dfrac45\right) \cdot \left(-\dfrac35\right) + \left(\left(-\dfrac45\right)^2-\left(\dfrac35\right)\right)^2 \cdot \dfrac45 \\ & = \dfrac{72}{125} + \left(\dfrac{16}{25}-\dfrac{9}{25}\right) \cdot \dfrac45 \\ & = \dfrac{72}{125} + \dfrac{28}{125} \\ & = \dfrac{100}{125} = \dfrac45 \end{aligned}$$Therefore, the value of $\boxed{\sin (2A+B) = \dfrac45}.$
(Answer B)

Use the sum and difference identities.
$$\boxed{\cos A-\cos B = -2 \sin \dfrac12(A+B) \sin \dfrac12(A-B)}$$Thus, we can write
$$\begin{aligned} \cos 265^{\circ}-\cos 95^{\circ} & = -2 \sin \dfrac12(265^{\circ}+95^{\circ}) \sin \dfrac12(265^{\circ}-95^{\circ}) \\ & = -2 \sin \dfrac12(360^{\circ}) \sin \dfrac12(170^{\circ}) \\ & = -2 \sin 180^{\circ} \sin 85^{\circ} \\ & = -2 \cdot 0 \cdot \sin 85^{\circ} \\ & = 0 \end{aligned}$$Therefore, the value of $\boxed{\cos 265^{\circ}-\cos 95^{\circ} = 0}.$
(Answer C)

Use the sum and difference identities.
$$\boxed{\sin A-\sin B = 2 \cos \dfrac12(A+B) \sin \dfrac12(A-B)}$$Thus, we can write
$$\begin{aligned} \sin 75^{\circ}-\sin 165^{\circ} & = 2 \cos \dfrac12(75^{\circ}+165^{\circ}) \sin \dfrac12(75^{\circ}-165^{\circ}) \\ & = 2 \cos \dfrac12(240^{\circ}) \sin \dfrac12(-90^{\circ}) \\ & = 2 \cos 120^{\circ} (-\sin 45^{\circ}) \\ & = 2 \cdot \left(-\dfrac12\right) \cdot \left(-\dfrac12\sqrt2\right) \\ & = \dfrac12\sqrt2 \end{aligned}$$Therefore, the value of $\boxed{\sin 75^{\circ}-\sin 165^{\circ} = \dfrac12\sqrt2}.$
(Answer D)

It is known that $\cos x = \dfrac35$ with $x$ located in the first quadrant.
Known: $\text{adjacent} = 3$ and $\text{hypotenuse} = 5$.
Using the right triangle approach and the Pythagorean theorem, we obtain
$\text{opposite} = \sqrt{5^2-3^2} = 4.$
From here, we obtain
$\sin x = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac45.$
Next, use the following trigonometric identities.
$$\boxed{\begin{aligned} \sin A + \sin B & = 2 \sin \dfrac12(A+B) \cos \dfrac12(A-B) \\ \sin 2x & = 2 \sin x \cos x \end{aligned}}$$We will obtain
$$\begin{aligned} \sin 3x + \sin x & = 2 \sin \dfrac12(3x+x) \cos \dfrac12(3x-x) \\ & = 2 \sin 2x \cos x \\ & = 2(2 \sin x \cos x) \cos x \\ & = 4 \sin x \cos^2 x \\ & = 4 \cdot \dfrac45 \cdot \left(\dfrac35\right)^2 \\ & = \dfrac{144}{125} \end{aligned}$$Therefore, the value of $\boxed{\sin 3x + \sin x = \dfrac{144}{125}}.$
(Answer E)

Use the following trigonometric identities.
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 \\ \cos (A-B) &= \cos A \cos B + \sin A \sin B \end{aligned}}$$It is known that $\sin A + \sin B = 1.$
Square both sides,
$\begin{aligned} (\sin A + \sin B)^2 & = 1^2 \\ \color{blue}{\sin^2 A + 2 \sin A \sin B + \sin^2 B} & \color{blue}{= 1} \end{aligned}$
It is known that $\cos A+ \cos B = \dfrac{\sqrt5}{\sqrt3}.$
Square both sides,
$$\begin{aligned} (\cos A + \cos B)^2 & = \left(\dfrac{\sqrt5}{\sqrt3}\right)^2 \\ \color{red}{\cos^2 A + 2 \cos A \cos B + \cos^2 B} & \color{red}{=\dfrac53}\end{aligned}$$Add the two equations marked in blue and red above.
$$\begin{aligned} (\sin^2 A + \cos^2 A) + 2 \sin A \sin B + 2 \cos A \cos B) + (\sin^2 B + \cos^2 B) & = 1+\dfrac53 \\ \cancel{1} + 2(\cos A \cos B + \sin A \sin B) + 1 & = \cancel{1}+\dfrac53 \\ 2(\cos A \cos B + \sin A \sin B) & = \dfrac23 \\ \cos A \cos B + \sin A \sin B & = \dfrac13 \\ \cos (A-B) & = \dfrac13 \end{aligned}$$Therefore, the value of $\boxed{\cos(A-B)=\dfrac13}.$
(Answer E)

It is known that $x-y = \dfrac{\pi}{6}$ and $\color{red}{\tan x = 3 \tan y}$ where $x, y$ are acute.
Using the tangent difference identity, we obtain
$\begin{aligned} \tan (x-y) & = \dfrac{\color{red}{\tan x}-\tan y}{1+\color{red}{\tan x} \tan y} \\ \tan \dfrac{\pi}{6} &= \dfrac{\color{red}{3 \tan y}-\tan y}{1+\color{red}{3 \tan y} \tan y} \\ \dfrac{1}{\sqrt3} & = \dfrac{2 \tan y}{1 + 3 \tan^2 y} \\ 1+3 \tan^2 y & = 2\sqrt3 \tan y \\ (\sqrt3 \tan y-1)^2 & = 0 \\ \sqrt3 \tan y-1 & = 0 \\ \tan y & = \dfrac{1}{\sqrt3} \\ \Rightarrow y & = \dfrac{\pi}{6} \end{aligned}$
Because $x-y=\dfrac{\pi}{6}$ and $y=\dfrac{\pi}{6}$, it means that $x=\dfrac{\pi}{3}$.
Therefore, the value of $\boxed{x+y=\dfrac{\pi}{3}+\dfrac{\pi}{6} = \dfrac{\pi}{2}}.$
(Answer A)

It is known:
$\begin{aligned} \sin \alpha & = \dfrac35 \\ \cos \beta & = \dfrac{12}{13} \\ \alpha, \beta~&\text{are acute}. \end{aligned}$
The sine value for angle $\beta$ and the cosine value for angle $\alpha$ will be positive because both $\alpha$ and $\beta$ are in the first quadrant.
Notice that
$\begin{aligned} \cos \alpha & = + \sqrt{1-\sin^2 \alpha} \\ & = \sqrt{1-\left(\dfrac35\right)^2} \\ & = \sqrt{1-\dfrac{9}{25}} \\ & = \sqrt{\dfrac{16}{25}} = \dfrac45 \end{aligned}$
and
$\begin{aligned} \sin \beta & = + \sqrt{1-\cos^2 \beta} \\ & = \sqrt{1-\left(\dfrac{12}{13}\right)^2} \\ & = \sqrt{1-\dfrac{144}{169}} \\ & = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13} \end{aligned}$
Using the sine angle sum identity, we obtain
$\begin{aligned} \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ & = \dfrac35 \cdot \dfrac{12}{13} + \dfrac45 \cdot \dfrac{5}{13} \\ & = \dfrac{36}{65}+\dfrac{20}{65} = \dfrac{56}{65} \end{aligned}$
Therefore, the value of $\boxed{\sin (\alpha + \beta) = \dfrac{56}{65}}.$
(Answer A)

It is known that $\cos x = \dfrac{12}{13}$.
Using the Pythagorean identity, we obtain
$\begin{aligned} \sin x & = \sqrt{1-\cos^2 x} \\ & = \sqrt{1-\left(\dfrac{12}{13}\right)^2} \\ & = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13} \end{aligned}$
Notice that $\tan \dfrac12x$ can be determined using the half-angle identity.
$\begin{aligned} \tan \dfrac12x & = \dfrac{1-\cos x}{\sin x} \\ & = \dfrac{1-\dfrac{12}{13}}{\dfrac{5}{13}} \\ & = \dfrac{\dfrac{1}{\cancel{13}}}{\dfrac{5}{\cancel{13}}} = \dfrac15 \end{aligned}$
Therefore, the value of $\boxed{\tan \dfrac12x = \dfrac15}.$
(Answer B)

Use the following double-angle identities.
$\boxed{\begin{aligned} \cos 2A & = 1-2 \sin^2 A \\ \sin 2A & = 2 \sin A \cos A \end{aligned}}$
We will obtain
$$\begin{aligned} \dfrac{1+\cos 2A}{\sin 2A} & = \dfrac{1 + (1-2 \sin^2 A)}{2 \sin A \cos A} \\ & = \dfrac{2-2 \sin^2 A}{2 \sin A \cos A} \\ & = \dfrac{2(1 -\sin^2 A)}{2 \sin A \cos A} \\ & = \dfrac{2 \cos^2 A}{2 \sin A \cos A} \\ & = \dfrac{\cos A}{\sin A} = \cot A \end{aligned}$$Therefore, another form of $\dfrac{1+\cos 2A}{\sin 2A}$ is $\boxed{\cot A}.$
(Answer C)

It is known:
$\begin{aligned} \tan \alpha & = 1 \\ \tan \beta & = \dfrac{1}{3} \\ \alpha, \beta~&\text{acute}. \end{aligned}$
The sine and cosine values for angle $\beta$ and angle $\alpha$ will be positive because both $\alpha, \beta$ are in the first quadrant.
Because $\tan \alpha = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{1}{1}$, then
$\begin{aligned} \sin \alpha & = \dfrac{ \text{opposite}}{\text{hypotenuse}} = + \dfrac{1}{\sqrt{1+1}} = \dfrac{1}{\sqrt2} \\ \cos \alpha & = \dfrac{ \text{adjacent}}{\text{hypotenuse}} = + \dfrac{1}{\sqrt{1+1}} = \dfrac{1}{\sqrt2} \end{aligned}$
Because $\tan \beta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{1}{3}$, then
$\begin{aligned} \sin \beta & = \dfrac{ \text{opposite}}{\text{hypotenuse}} = + \dfrac{1}{\sqrt{1+3^2}} = \dfrac{1}{\sqrt{10}} \\ \cos \beta & = \dfrac{ \text{adjacent}}{\text{hypotenuse}} = +\dfrac{3}{\sqrt{1+3^2}} = \dfrac{3}{\sqrt{10}} \end{aligned}$
Using the sine difference identity, we obtain
$$\begin{aligned} \sin (\alpha- \beta) & = \sin \alpha \cos \beta-\cos \alpha \sin \beta \\ & = \dfrac{1}{\sqrt{2}} \cdot \dfrac{3}{\sqrt{10}}-\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{10}} \\ & = \dfrac{3}{\sqrt{20}}-\dfrac{1}{\sqrt{20}} = \dfrac{2}{\sqrt{20}} = \dfrac15\sqrt5 \end{aligned}$$Therefore, the value of $\boxed{\sin (\alpha-\beta) = \dfrac15\sqrt5}.$
(Answer B)

Notice that $\sin (\alpha-\beta) = \dfrac16$ can be rewritten in another form using the sine difference identity, namely
$\color{red}{\sin \alpha \cos \beta-\cos \alpha \sin \beta = \dfrac16}.$
From the equation $\tan \alpha-\tan \beta = \dfrac15$, we obtain
$\begin{aligned} \dfrac{\sin \alpha}{\cos \alpha}-\dfrac{\sin \beta}{\cos \beta} & = \dfrac15 \\ \dfrac{\color{red}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}}{\cos \alpha \cos \beta} & = \dfrac15 \\ \dfrac{\color{red}{\frac16}}{\cos \alpha \cos \beta} & = \dfrac15 \\ \cos \alpha \cos \beta & = \dfrac56 \end{aligned}$
Therefore, the value of $\boxed{\cos \alpha \cos \beta=\dfrac56}.$
(Answer D)

Given $\sin x + \cos x = -\dfrac15$.
Square both sides to obtain
$$\begin{aligned} (\sin x + \cos x)^2 & = \left(-\dfrac15\right)^2 \\ \color{red}{\sin^2 x} + \color{blue}{2 \sin x \cos x} + \color{red}{\cos^2 x} & = \dfrac{1}{25} \\ 1 + \sin 2x & = \dfrac{1}{25} \\ \sin 2x & = \dfrac{1}{25}-1 \\ &= -\dfrac{24}{25} \end{aligned}$$Note:
$\boxed{\begin{aligned} \color{red}{\sin^2 x + \cos^2 x} & = \color{red}{1} \\ \color{blue}{2 \sin x \cos x} & = \color{blue}{\sin 2x} \end{aligned}}$
Therefore, the value of $\boxed{\sin 2x = -\dfrac{24}{25}}.$
(Answer A)

From the equation
$\color{blue}{\sin \theta + \cos \theta = \dfrac12}$,
square both sides to obtain
$\begin{aligned} (\sin \theta + \cos \theta)^2 & = (\dfrac12)^2 \\ \color{red}{\sin^2 \theta} + 2 \sin \theta \cos \theta + \color{red}{\cos^2 \theta} & = \dfrac14 \\ \color{red}{1} + 2 \sin \theta \cos \theta & = \dfrac14 \\ 2 \sin \theta \cos \theta & = -\dfrac34 \\ \sin \theta \cos \theta & = -\dfrac38 \end{aligned}$
Use the factorization formula:
$a^3+b^3 = (a+b)^3-3ab(a+b)$
For $a = \sin \theta$ and $b = \cos \theta$, we obtain
$$\begin{aligned} \sin^3 \theta + \cos^3 \theta & = (\color{blue}{\sin \theta + \cos \theta})^3-3 \sin \theta \cos \theta(\color{blue}{\sin \theta + \cos \theta}) \\ & = \left(\dfrac12\right)^3-3\left(-\dfrac38\right)\left(\dfrac12\right) \\ & = \dfrac18 + \dfrac{9}{16} = \dfrac{11}{16} \end{aligned}$$Therefore, the value of $\boxed{\sin^3 \theta + \cos^3 \theta = \dfrac{11}{16}}.$
(Answer E)

First, convert the matrix equation above into ordinary algebraic equations by performing matrix multiplication.
We will obtain two equations, namely
$$\begin{cases} \tan x \cos^2 x + \sin x \cos x & = \dfrac{a}{2} && (\cdots 1) \\ \cos^2 x + \tan x \sin x \cos x & = \dfrac{b}{2} && (\cdots 2) \end{cases}$$Observe equation $(2)$.
Since $\tan x = \dfrac{\sin x}{\cos x}$ (ratio identity) and $\cos^2 x = 1-\sin^2 x$ (Pythagorean Identity), then we obtain
$$\begin{aligned} (1-\sin^2 x)+\dfrac{\sin x}{\cancel{\cos x}} \cdot \sin x \cdot \cancel{\cos x} & = \dfrac{b}{2} \\ 1-\sin^2 x + \sin^2 x & = \dfrac{b}{2} \\ 1 & = \dfrac{b}{2} \\ b & = 2 \end{aligned}$$Since it is known that $b=2a$ and we obtain $b=2$, then $a=1$.
Substitute $a=1$ into equation $(1)$.
$$\begin{aligned} \tan x \cos^2 x + \sin x \cos x & = \dfrac12 \\ \dfrac{\sin x}{\cancel{\cos x}} \cdot \cancelto{\cos x}{\cos^2 x} + \sin x \cos x & = \dfrac12 \\ \sin x \cos x + \sin x \cos x & = \dfrac12 \\ \color{red}{2 \sin x \cos x} & = \dfrac12 \\ \color{red}{\sin 2x} & = \dfrac12 \end{aligned}$$It is known that $0 \leq x \leq \pi.$
Sine equals $\dfrac12$ when the angle is $30^{\circ}$ or $150^{\circ}$. Therefore, we write
$\sin 2x = \sin 30^{\circ}$ which means $x = 15^{\circ}$ and
$\sin 2x = \sin 150^{\circ}$ which means $x = 75^{\circ}.$
Therefore, the value of $x$ that satisfies based on the available answer choices is $\boxed{15^{\circ}}.$
(Answer E)

Given $\sin \alpha = \dfrac{\sqrt3}{3} = \dfrac{\text{opposite}}{\text{hypotenuse}}.$
Using the Pythagorean Theorem, we obtain
$\text{adjacent} = \sqrt{(3)^2-(\sqrt3)^2} = \sqrt6$
Therefore, we get
$\begin{aligned} \cos \alpha & = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{\sqrt6}{3} \\ \cot \alpha & = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{\sqrt6}{\sqrt3} = \sqrt2 \end{aligned}$
Notice that $\tan \left(\dfrac12 \pi-\alpha\right) = \cot \alpha$.
Thus,
$$\begin{aligned} \tan \left(\dfrac12 \pi-\alpha\right)+3 \cos \alpha & = \cot \alpha + 3 \cos \alpha \\ & = \sqrt2 + \cancel{3} \cdot \dfrac{\sqrt6}{\cancel{3}} \\ & = \sqrt6 + \sqrt2 \end{aligned}$$Therefore, the value of $\boxed{\tan \left(\dfrac12 \pi-\alpha\right)+3 \cos \alpha = \sqrt6 + \sqrt2}.$
(Answer C)

Using the double-angle identity:
$\boxed{\sin 2A = 2 \sin A \cos A}$
it is obtained that
$\begin{aligned} & 2 \cos \left(\dfrac14 \pi + x\right) \sin \left(\dfrac14 \pi + x\right) \\ & = \sin 2\left(\dfrac14 \pi + x\right) \\ & = \sin \left(\dfrac12 \pi + 2x\right) \\ & = \cos 2x \end{aligned}$
Therefore, the value of $$\boxed{2 \cos \left(\dfrac14 \pi + x\right) \sin \left(\dfrac14 \pi + x\right) = \cos 2x}.$$(Answer E)

Use the sum identity of sine functions:
$$\boxed{\sin A + \sin B = 2 \sin \dfrac{A+B}{2} \cos \dfrac{A+B}{2}}$$We will obtain
$$\begin{aligned} & \color{red}{\sin 160^{\circ} + \sin 140^{\circ}}-\cos 10^{\circ} \\ & = 2 \sin \dfrac{160^{\circ}+140^{\circ}}{2} \cos \dfrac{160^{\circ}-140^{\circ}}{2}-\cos 10^{\circ} \\ & = 2 \sin 150^{\circ} \cos 10^{\circ}-\cos 10^{\circ} \\ & = 2 \cdot \dfrac12 \cdot \cos 10^{\circ}-\cos 10^{\circ} = 0 \end{aligned}$$Therefore, the value of $\boxed{\sin 160^{\circ} + \sin 140^{\circ}-\cos 10^{\circ} = 0}.$
(Answer C)

Use the Pythagorean identity and the angle relation identity in the first quadrant below.
$\boxed{\begin{aligned} \cos^2 x & = 1-\sin^2 x \\ \sin x & = \cos (90^{\circ}-x) \end{aligned}}$
We will obtain
$$\begin{aligned} & \cos^2 30^{\circ} + \cos^2 40^{\circ}+\cos^2 50^{\circ} +\cos^2 60^{\circ} \\ & = (1-\cancel{\sin^2 30^{\circ}}) + (1-\bcancel{\sin^2 40^{\circ}}) + \bcancel{\sin^2 40^{\circ}} + \cancel{\sin^2 30^{\circ}} \\ & = 1+1=2 \end{aligned}$$Therefore, the value of $$\boxed{\cos^2 30^{\circ} + \cos^2 40^{\circ}+\cos^2 50^{\circ} +\cos^2 60^{\circ}=2}.$$(Answer A)

Given $\triangle ABC$ (right triangle). It is known that $\sin A \sin B = 0,\!5.$
It is also known that the right angle is at point $A$, which means it can be written as
$\sin 90^{\circ} \sin B = 0,\!5.$
Since $\sin 90^{\circ} = 1$, then $\sin B = 0,\!5.$
Using the sine and cosine angle relation, it is obtained
$\begin{aligned} \cos (A+B) & = \cos (90^{\circ}+B) \\ & = -\sin B = -0,\!5 \end{aligned}$
Therefore, the value of $\boxed{\cos (A+B)=-0,\!5}.$
(Answer D)

Notice that $\cos A \cos B = \dfrac13$. If $A = 90^{\circ}$ or $B = 90^{\circ}$, then the equation does not hold because $\cos 90^{\circ} = 0$. This means the right angle is at $C$, written as $\angle C = 90^{\circ}$.
This also means that $B = 90^{\circ}-A.$
Thus, it is obtained
$\begin{aligned} \cos A \cos B & = \dfrac13 \\ \cos A \cos (90^{\circ}-A) & = \dfrac13 \\ \cos A \sin A & = \dfrac13 \\ \text{Multiply}~2~\text{on both sides} \\ 2 \cos A \sin A & = \dfrac23 \\ \sin 2A & = \dfrac23 \end{aligned}$
Note: Remember that
$\boxed{\begin{aligned} \cos (90^{\circ}-A) & = \sin A \\ 2 \sin A \cos A & = \sin 2A \end{aligned}}$
Using the right triangle approach as shown in the figure,
it is obtained that $\boxed{\cos 2A = \dfrac{\text{adj}}{\text{hyp}} = \dfrac13\sqrt5}.$
(Answer E)

Notice that by multiplying both sides by $2$, we obtain
$\begin{aligned} \dfrac12x+y =\dfrac{\pi}{4} \Rightarrow x + 2y & = \dfrac{\pi}{2} \\ x & = \dfrac{\pi}{2}-2y \end{aligned}$
Next, use the double-angle identity for tangent.
$\boxed{\begin{aligned} \tan (90^{\circ}-\theta) & = \cot \theta \\ \tan 2\theta & = \dfrac{2 \tan \theta}{1-\tan^2 \theta} \end{aligned}}$
For cotangent, simply swap the numerator and denominator positions.
Thus,
$\begin{aligned} \tan x & = \tan \left(\dfrac{\pi}{2}-2y\right) \\ & = \cot 2y \\ & = \dfrac{1-\tan^2 y}{2 \tan y} \end{aligned}$
Therefore, we obtain $\boxed{\tan x = \dfrac{1-\tan^2 y}{2 \tan y}}.$
(Answer A)

Given $\sin (3x+2y) = \dfrac13$ and $\cos (3x-4y) = \dfrac34$.
Let $\alpha = 3x + 2y$ and $\beta = 3x-4y$ so that
$$\begin{aligned} 6y & = (3x+2y)-(3x-4y) = \alpha-\beta \\ 9x &= 2(3x+2y)+(3x-4y) = 2\alpha+\beta \end{aligned}$$

Notice that $\sin \alpha = \dfrac13$ so by using the right triangle approach, we obtain $\cos \alpha = \dfrac{2\sqrt2}{3}$. Likewise, for $\cos \beta = \dfrac34$, we obtain $\sin \beta = \dfrac{\sqrt7}{4}.$
Next, we obtain
$\begin{aligned} \sin 6y & = \sin (\alpha-\beta) \\ & = \sin \alpha \cos \beta+\cos \alpha \sin \beta \\ & = \dfrac13 \cdot \dfrac34-\dfrac{\sqrt7}{4} \cdot \dfrac{2\sqrt2}{3} \\ & = \dfrac{3-2\sqrt{14}}{12} \end{aligned}$
and
$$\begin{aligned} \cos 9x & = \cos (2\alpha+\beta) \\ & = \cos 2\alpha \cos \beta-\sin 2\alpha \sin \beta \\ & = (\cos^2 \alpha-\sin^2 \alpha) \cos \beta-2 \sin \alpha \cos \alpha \sin \beta \\ & = \left(\left(\dfrac{2\sqrt2}{3}\right)^2-\left(\dfrac13\right)^2\right) \cdot \dfrac34-\cancel{2} \cdot \dfrac13 \cdot \dfrac{2\sqrt2}{3} \cdot \dfrac{\sqrt7}{\cancelto{2}{4}} \\ & = \dfrac{7}{12}-\dfrac{2\sqrt{14}}{18} = \dfrac{21-4\sqrt{14}}{36} \end{aligned}$$Thus, we get
$\begin{aligned} \dfrac{\sin 6y}{\cos 9x} & = \dfrac{3-2\sqrt{14}}{\cancel{12}} \times \dfrac{\cancelto{3}{36}}{21-4\sqrt{14}} \\ & = \dfrac{9-6\sqrt{14}}{21-4\sqrt{14}} \end{aligned}$
Therefore, the value of $\dfrac{\sin 6y}{\cos 9x}$ is $\boxed{\dfrac{9-6\sqrt{14}}{21-4\sqrt{14}}}.$
(Answer A)

Use the following trigonometric identities.
$$\boxed{\begin{aligned} 2 \cos x \cos y & = \cos (x+y) + \cos (x-y) \\ \cos x + \cos y & = 2 \cos \dfrac12(x+y) \cos \dfrac12(x-y) \end{aligned}}$$For that, we can obtain
$$\begin{aligned} & 20 \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} \\ & = 10(2 \cos 20^{\circ} \cos 40^{\circ}) \cos 80^{\circ} \\ & = 10(\cos 60^{\circ} + \cos (-20^{\circ})) \cos 80^{\circ} \\ & = 10 \cos 60^{\circ} \cos 80^{\circ} + 10 \cos 20^{\circ} \cos 80^{\circ} \\ & = 10 \left(\dfrac12\right) \cos 80^{\circ} + 5(2 \cos 20^{\circ} \cos 80^{\circ}) \\ & = 5 \cos 80^{\circ} + 5(\cos 100^{\circ} + \cos (-60^{\circ}) \\ & = 5(\cos 80^{\circ} + \cos 100^{\circ}) + \dfrac52 \\ & = 5(2 \cos \dfrac12(80^{\circ}+100^{\circ}) \cos \dfrac12(80^{\circ}-100^{\circ}) + \dfrac52 \\ & = 5(2 \cos 90^{\circ} \cos 10^{\circ}) + \dfrac52 \\ & = 5(2(0) \cos 10^{\circ}) + \dfrac52 \\ & = \dfrac52 \end{aligned}$$Therefore, the value of $\boxed{20 \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} = \dfrac52}.$
(Answer C)

Given $\sin (40+x)^{\circ} = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{a}{1}$ with $(40^{\circ}+x)$ located in the first quadrant.
Based on the Pythagorean formula, we obtain
$\text{adjacent} = \sqrt{1^2-a^2} = \sqrt{1-a^2}$
thus
$\begin{aligned} \cos (40^{\circ} + x) & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ & = \dfrac{\sqrt{1-a^2}}{1} \\ & = \sqrt{1-a^2} \end{aligned}$
Therefore,
$$\begin{aligned} \cos (70^{\circ}+x) & = \color{red}{\cos ((40^{\circ}+x)+30^{\circ})} \\ & = \cos (40^{\circ}+x) \cos 30^{\circ}-\sin (40^{\circ}+x) \sin 30^{\circ} \\ & = \sqrt{1-a^2} \cdot \dfrac12\sqrt3-a \cdot \dfrac12 \\ & = \dfrac12\sqrt{3(1-a^2)}-\dfrac12a \\ & = \dfrac{\sqrt{3(1-a^2)}-a}{2} \end{aligned}$$Note: use the sum and difference angle formulas.
Therefore, the value of $\boxed{\cos (70^{\circ}+x) = \dfrac{\sqrt{3(1-a^2)}-a}{2}}.$
(Answer B)

Given $\cos (x+15^{\circ}) = a$ with $0^{\circ} \leq x \leq 30^{\circ}.$
Notice that
$\begin{aligned} \color{red}{\cos (2x+30^{\circ})} & = \cos 2(x+15^{\circ}) \\ & = 2 \cos^2 (x+15^{\circ})-1 \\ & = \color{red}{2a^2-1} \end{aligned}$
Therefore,
$$\begin{aligned} \color{blue}{\sin (2x+30^{\circ})} & = \sqrt{1-\cos^2 (2x+30^{\circ})} \\ & = \sqrt{1-(2a^2-1)^2} \\ & = \sqrt{1-(4a^4-4a^2+1)} \\ & = \sqrt{4a^2(1-a^2)} \\ & = \color{blue}{2a\sqrt{1-a^2}} \end{aligned}$$Next, using the cosine sum identity, we obtain
$$\begin{aligned} \cos (2x+60^{\circ}) & = \cos ((2x+30^{\circ})+30^{\circ}) \\ & = \color{red}{\cos (2x+30^{\circ})} \cdot \cos 30^{\circ}-\color{blue}{\sin (2x+30^{\circ})} \cdot \sin 30^{\circ} \\ & = \color{red}{(2a^2-1)} \cdot \dfrac12\sqrt3-\color{blue}{2a\sqrt{1-a^2}} \cdot \dfrac12 \\ & = \dfrac12\sqrt3(2a^2-1)-a\sqrt{1-a^2} \end{aligned}$$Therefore, the value of
$$\boxed{\cos (2x+60^{\circ}) = \dfrac12\sqrt3(2a^2-1)-a\sqrt{1-a^2}}.$$(Answer B)

Observe the following sketch.
From the figure, it is known that
$\begin{aligned} |AB| & = 3 \\ |AC| & = 5 \\ |BF| & = 2 \\ |CD| & = 1 \end{aligned}$
Note: A notation such as $|AB|$ means the length of $AB$.
Using the Pythagorean Theorem, we obtain
$\begin{aligned} |AF| & = \sqrt{|AB|^2+|BF|^2} \\ & = \sqrt{3^2+2^2} = \sqrt{13} \\ |AD| & = \sqrt{|AC|^2+|CD|^2} \\ & = \sqrt{5^2+1^2} = \sqrt{26} \end{aligned}$
so that the following trigonometric ratios apply
$\begin{aligned} \sin \alpha & = \dfrac{|CD|}{|AD|} = \dfrac{1}{\sqrt{26}} \\ \cos \alpha & = \dfrac{|AC|}{|AD|} = \dfrac{5}{\sqrt{26}} \\ \sin \beta & = \dfrac{|AB|}{|AF|} = \dfrac{3}{\sqrt{13}} \\ \cos \beta & = \dfrac{|BF|}{|AF|} = \dfrac{2}{\sqrt{13}} \end{aligned}$
Using the sine difference identity, we obtain
$$\begin{aligned} \sin (\beta-\cos \alpha) & = \sin \beta \cos \alpha-\sin \alpha \cos \beta \\ & = \dfrac{3}{\sqrt{13}} \cdot \dfrac{5}{\sqrt{26}}- \dfrac{1}{\sqrt{26}} \cdot \dfrac{2}{\sqrt{13}} \\ & = \dfrac{15}{\sqrt{13^2 \cdot 2}}-\dfrac{2}{\sqrt{13^2 \cdot 2}} \\ & = \dfrac{13}{13\sqrt2} = \dfrac12\sqrt2 \end{aligned}$$Since $\alpha, \beta$ are acute, then $(\alpha-\beta)$ must also be acute, so the possible value for $(\beta-\alpha)$ is $45^{\circ}.$
(Answer B)

Substitute $\cos^2 \theta = 1-\sin^2 \theta$ into the given equation to obtain
$\begin{aligned} & \sin^2 \theta + (1-\sin^2 \theta)^2 \\ & = \sin^2 \theta + (1-2 \sin^2 \theta + \sin^4 \theta) \\ & = 1- \sin^2 \theta + \sin^4 \theta \\ & = 1-(\sin^2 \theta)(1-\sin^2 \theta) \\ & = 1-\sin^2 \theta \cos^2 \theta \\ & = 1-(\sin \theta \cos \theta)^2 \\ & = 1-\left(\dfrac12 \sin 2\theta\right)^2 = 1-\dfrac14 \sin^2 2\theta \end{aligned}$
Note: $2 \sin x \cos x = \sin 2x$.
Since the value of $\sin \theta$ lies in the interval $-1 \leq \sin \theta \leq 1$, then the following applies
$\begin{array}{rcccl} -1 & \leq & \sin 2\theta & \leq & 1 \\ 0 & \leq & \sin^2 2\theta & \leq & 1 \\ 0 & \leq & \dfrac14 \sin^2 2\theta & \leq & 1 \\ -\dfrac14 & \leq & -\dfrac14 \sin^2 2\theta & \leq & 0 \\ 1-\dfrac14 & \leq & 1-\dfrac14 \sin^2 2\theta & \leq & 1+0 \\ \dfrac34 & \leq & \dfrac14 \sin^2 2\theta & \leq & 1 \end{array}$
Therefore, for all real values of $\theta$, the following holds $\boxed{\dfrac34 \leq \alpha \leq 1}.$
(Answer D)

Answer a)
Based on the double-angle identity:
$\boxed{\cos 2x = 1-2 \sin^2 x}$
we obtain
$\begin{aligned} 1-2 \sin^2 22,5^{\circ} & = \cos 2(22,5^{\circ}) \\ & = \cos 45^{\circ} \\ & = \dfrac12\sqrt2 \end{aligned}$

Answer b)
Based on the product identity:
$$\boxed{2 \sin A \cos B = \sin (A+B)+\sin (A-B)}$$we obtain
$$\begin{aligned} & 2 \cos 97,5^{\circ} \sin 52,5^{\circ} \\ & = \sin (52,5^{\circ}+97,5^{\circ}) + \sin (52,5^{\circ}-97,5^{\circ}) \\ & = \sin (150^{\circ}) + \sin (-45^{\circ}) \\ & = \sin 150^{\circ}- \sin 45^{\circ} \\ & = \dfrac12-\dfrac12\sqrt2 \\ & = \dfrac12(1-\sqrt2) \end{aligned}$$

Answer c)
Based on the product identity:
$$\boxed{2 \sin A \cos B = \sin (A+B)+\sin (A-B)}$$we obtain
$\begin{aligned} & 6 \sin 165^{\circ} \cos 15^{\circ} \\ & = 3(2 \sin 165^{\circ} \cos 15^{\circ}) \\ & = 3(\sin (165^{\circ} + 15^{\circ}) + \sin (165^{\circ} -15^{\circ}) \\ & = 3(\sin 180^{\circ} + \sin 150^{\circ}) \\ & = 3\left(0 + \dfrac12\right) = \dfrac32 \end{aligned}$

Answer d)
Based on the difference identities of sine and cosine functions:
$$\boxed{\begin{aligned} \sin A-\sin B & = 2 \cos \dfrac{A+B}{2} \sin \dfrac{A+B}{2} \\ \cos A-\cos B & = -2 \sin \dfrac{A+B}{2} \sin \dfrac{A+B}{2} \end{aligned}}$$we obtain
$\begin{aligned} & \dfrac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}} \\ & = \dfrac{-2 \sin \dfrac{75^{\circ}+15^{\circ}}{2} \cos \dfrac{75^{\circ}-15^{\circ}}{2}}{2 \cos \dfrac{75^{\circ}+15^{\circ}}{2} \sin \dfrac{75^{\circ}-15^{\circ}}{2}} \\ & = \dfrac{-\cancel{2} \sin 45^{\circ} \bcancel{\sin 30^{\circ}}}{\cancel{2} \cos 45^{\circ} \bcancel{\sin 30^{\circ}}} \\ & = \dfrac{-\cancel{\dfrac12\sqrt2}}{\cancel{\dfrac12\sqrt2}} \\ & = -1 \end{aligned}$

Based on the sine angle-sum identity:
$\boxed{\sin (x+y) = \sin x \cos y + \cos x \sin y}$
we obtain
$$\begin{aligned} \sin \left(A+\dfrac{\pi}{3}\right) & = \sin A \\ \sin A \cos \dfrac{\pi}{3} + \cos A \sin \dfrac{\pi}{3} & = \sin A \\ \sin A \cdot \dfrac12 + \cos A \cdot \dfrac12\sqrt3 & = \sin A \\ \dfrac12\sin A-\sin A & = -\dfrac12\sqrt3 \cos A \\ \bcancel{-\dfrac12} \sin A & = \bcancel{-\dfrac12}\sqrt3 \cos A \\ \sin A & = \sqrt3 \cos A \\ \dfrac{\sin A}{\cos A} = \tan A & = \sqrt3 \end{aligned}$$Therefore, it is proven that $\tan A = \sqrt3$.

Based on the cosine sum and difference identities:
$\boxed{\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y}$
we obtain
$$\begin{aligned} 3 \cos \left(A+\dfrac{\pi}{4}\right) & = \cos \left(A-\dfrac{\pi}{4}\right) \\ 3 \left(\cos A \cos \dfrac{\pi}{4}-\sin A \sin \dfrac{\pi}{4}\right) & = \cos A \cos \dfrac{\pi}{4}+\sin A \sin \dfrac{\pi}{4} \\ 3 \left(\cos A \cdot \dfrac12\sqrt2 -\sin A \cdot \dfrac12\sqrt2\right) & = \cos A \cdot \dfrac12\sqrt2 +\sin A \cdot \dfrac12\sqrt2 \\ \dfrac32\sqrt2 \cos A-\dfrac32\sqrt2 \sin A & = \dfrac12\sqrt2 \cos A+ \dfrac12\sqrt2 \sin A \\ \dfrac32\sqrt2 \cos A-\dfrac12\sqrt2 \cos A & = \dfrac12\sqrt2 \sin A+\dfrac32\sqrt2 \sin A \\ \bcancel{\sqrt2} \cos A & = 2 \cdot \bcancel{\sqrt2} \sin A \\ \cos A & = 2 \sin A \\ \dfrac{\sin A}{\cos A} = \tan A & = \dfrac12 \end{aligned}$$Therefore, it is proven that $\tan A = \dfrac12$.

Answer a)
The equation is given by
$$\cos x = \cos 100^{\circ} \cos 35^{\circ}-\sin 100^{\circ} \sin 35^{\circ}.$$Use the cosine angle-sum identity.
$\boxed{\cos (A+B) = \cos A \cos B-\sin A \sin B}$
For $A = 100^{\circ}$ and $B = 35^{\circ}$, we obtain
$\begin{aligned} & \cos 100^{\circ} \cos 35^{\circ}-\sin 100^{\circ} \sin 35^{\circ} \\ & = \cos (100^{\circ} + 35^{\circ}) \\ & = \cos 135^{\circ} = -\dfrac12\sqrt2 \end{aligned}$
Therefore, the value of $x = 135^{\circ}$. In addition, the value of $x$ can also be $225^{\circ}$ because $\cos 225^{\circ} = -\dfrac12\sqrt2$.

Answer b)
The equation is given by
$\sin x = \sin 115^{\circ} \cos 5^{\circ} + \cos 115^{\circ} \sin 5^{\circ}$
Use the sine angle-sum identity.
$$\boxed{\sin (A+B) = \sin A \cos B-\cos A \sin B}$$For $A = 115^{\circ}$ and $B = 5^{\circ}$, we obtain
$\begin{aligned} & \sin 115^{\circ} \cos 5^{\circ} + \cos 115^{\circ} \sin 5^{\circ} \\ & = \sin (115^{\circ} + 5^{\circ}) \\ & = \sin 120^{\circ} = \dfrac12\sqrt3 \end{aligned}$
Therefore, the value of $x = 120^{\circ}$. In addition, the value of $x$ can also be $60^{\circ}$ because $\sin 60^{\circ} = \dfrac12\sqrt3$.

Answer c)
The equation is given by
$\tan 3x = \dfrac{\tan \dfrac{7}{12}\pi-\tan \dfrac{\pi}{12}}{1+\tan \dfrac{7}{12}\pi \tan \dfrac{\pi}{12}}$
Use the tangent angle-difference identity.
$\boxed{\tan (A-B) = \dfrac{\tan A-\tan B}{1+\tan A \tan B}}$
For $A = \dfrac{7}{12}\pi$ and $B = \dfrac{\pi}{12}$, we obtain
$$\begin{aligned} \dfrac{\tan \dfrac{7}{12}\pi-\tan \dfrac{\pi}{12}}{1+\tan \dfrac{7}{12}\pi \tan \dfrac{\pi}{12}} & = \tan \left(\dfrac{7}{12}\pi-\dfrac{\pi}{12}\right) \\ & = \tan \dfrac12\pi = \tan 3x \end{aligned}$$Based on the basic trigonometric equation, we obtain the following form.
$\begin{aligned} 3x & = \dfrac12\pi + k \cdot \pi \\ x & = \dfrac16\pi + k \cdot \dfrac{\pi}{3} \end{aligned}$
For $k = 0, 1, 2, 3, 4, 5$, respectively, we obtain $x = \dfrac16\pi$, $x = \dfrac12\pi$, $x = \dfrac56\pi$, $x = \dfrac76\pi$, $\dfrac32\pi$, and $x = \dfrac{11}{6}\pi$.

Since $\sin A = \dfrac35$, then by using the Pythagorean identity, we obtain
$\begin{aligned} \cos a &= \sqrt{1-\sin^2 a} \\ & = \sqrt{1-\left(\dfrac35\right)^2} \\ & = \sqrt{1-\dfrac{9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac45 \end{aligned}$
Using the same principle, from $\cos b = \dfrac{5}{13}$, we obtain
$\begin{aligned} \sin b & = \sqrt{1-\cos^2 b} \\ & = \sqrt{1-\left(\dfrac{5}{13}\right)^2} \\ & = \sqrt{1-\dfrac{25}{169}} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13} \end{aligned}$
Notice that $a+b+c=180^{\circ}$ so that $c=180^{\circ}-(a+b)$. Therefore, we obtain
$\begin{aligned} \sin c & = \sin (180^{\circ}-(a+b)) \\ & = \sin (a+b) \\ & = \sin a \cos b + \cos a \sin b \\ & = \dfrac35 \cdot \dfrac{5}{13} + \dfrac45 \cdot \dfrac{12}{13} \\ & = \dfrac{15}{65}+\dfrac{48}{65} = \dfrac{63}{65} \end{aligned}$
Therefore, the value of $\boxed{\sin c = \dfrac{63}{65}}.$

We will determine the values of $\tan x$ and $\tan y$ first by observing the figure above.
Based on the Pythagorean theorem, the length of $KM$ is formulated by
$\begin{aligned} KM & = \sqrt{KL^2+LM^2} \\ & = \sqrt{24^2+7^2} \\ & = \sqrt{625} = 25~\text{cm} \end{aligned}$
Also based on the Pythagorean theorem, the length of $KN$ is formulated by
$\begin{aligned} KN & = \sqrt{KM^2-MN^2} \\ & = \sqrt{25^2-15^2} \\ & = \sqrt{400} = 20~\text{cm} \end{aligned}$
Thus,
$\tan x = \dfrac{LM}{KL} = \dfrac{7}{24}$
$\tan y = \dfrac{MN}{KN} = \dfrac{15}{20} = \dfrac34.$
Use the tangent angle-sum identity.
$\begin{aligned} \tan (x+y) & = \dfrac{\tan x + \tan y}{1-\tan x \tan y} \\ & = \dfrac{\dfrac{7}{24} + \dfrac34}{1-\dfrac{7}{24} \cdot \dfrac34} \\ & = \dfrac{\dfrac{25}{24}}{1-\dfrac{21}{96}} \\ & = \dfrac{\cancel{25}}{\cancel{24}} \cdot \dfrac{\cancelto{4}{96}}{\cancelto{3}{75}} \\ & = \dfrac43 \end{aligned}$
Therefore, the value of $\boxed{\tan (x+y) = \dfrac{4}{3}}.$

To prove the statement above, we will use the tangent angle-sum identity.
$$\begin{aligned} \tan (x+y) & = \dfrac{\tan x + \tan y }{1-\tan x \tan y} \\ \tan \dfrac54\pi & = \dfrac{\tan x + \tan y }{1-\tan x \tan y} \\ 1 & = \dfrac{\tan x + \tan y }{1-\tan x \tan y} \\ 1-\tan x \tan y & = \tan x + \tan y \\ \tan x + \tan y + \tan x \tan y-1 & = 0 \\ (\tan x + 1)(\tan y +1)-2 & = 0 \\ (\tan x + 1)(\tan y +1) & = 2 \end{aligned}$$Therefore, it is proven that $\boxed{(\tan x + 1)(\tan y +1)=2}.$

Use the sine angle-sum identity.
$$\begin{aligned} \sin (x+60^{\circ}) & = \sin x \\ \sin x \cos 60^{\circ} + \cos x \sin 60^{\circ} & = \sin x \\ \dfrac12 \sin x + \dfrac12\sqrt3 \cos x & = \sin x \\ \cancel{\dfrac12}\sqrt3 \cos x & = \cancel{\dfrac12} \sin x \\ \sqrt3 \cos x & = \sin x \\ \dfrac{\sin x}{\cos x} = \tan x & = \sqrt3 \end{aligned}$$Therefore, it is proven that $\tan x = \sqrt3$.

Given
$\begin{aligned} p & = \sin x + \sin y \\ q & = \cos x + \cos y \end{aligned}$
It will be proven that $\dfrac12(p^2+q^2) = 1+\cos (x-y)$ by using the following trigonometric identities.
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 \\ \cos (x-y) & = \cos x \cos y + \sin x \sin y \end{aligned}}$$We will obtain
$$\begin{aligned} \dfrac12(p^2+q^2) & = \dfrac12((\sin x+\sin y)^2+(\cos x + \cos y)^2) \\ & = \dfrac12(\color{red}{\sin^2 x} + 2 \sin x \sin y + \color{blue}{\sin^2 y} + \color{red}{\cos^2 x} + 2 \cos x \cos y + \color{blue}{\cos^2 y}) \\ & = \dfrac12(2(\cos x \cos y + \sin x \sin y)+\color{red}{1}+\color{blue}{1}) \\ & = (\cos x \cos y + \sin x \sin y)+1 \\ & = 1+\cos (x-y) \end{aligned}$$Therefore, it is proven that $\dfrac12(p^2+q^2) = 1+\cos (x-y)$.

Answer a)
By using the following trigonometric identities:
$$\boxed{\begin{aligned} \sin (A \pm B) & = \sin A \cos B \pm \cos A \sin B \\ \sin^2 A + \cos^2 A & = 1 \end{aligned}}$$we obtain
$$\begin{aligned} \sin (x+y) \sin (x-y) & =(\sin x \cos y + \cos x \sin y)(\sin x \cos y-\cos x \sin y) \\ & = \sin^2 x \cos^2 y- \bcancel{\sin x \sin y \cos x \cos y} + \bcancel{\sin x \sin y \cos x \cos y} -\cos^2 x \sin^2 y \\ & = \sin^2 x \cos^2 y-\cos^2 x \sin^2 y \\ & = \sin^2 x(1-\sin^2 y)-(1-\sin^2 x) \sin^2 y \\ & = \sin^2 x-\cancel{\sin^2 x \sin^2 y}-\sin^2 y+\cancel{\sin^2 x \sin^2 y} \\ & = \sin^2 x-\sin^2 y \end{aligned}$$Therefore, it is proven that $\sin (x+y) \sin (x-y) = \sin^2 x-\sin^2 y$.

Answer b)
By using the following trigonometric identities:
$$\boxed{\begin{aligned} \sin (A \pm B) & = \sin A \cos B \pm \cos A \sin B \\ \sin^2 A + \cos^2 A & = 1 \end{aligned}}$$we obtain
$$\begin{aligned} \cos (x+y) \cos (x-y) & =(\cos x \cos y- \sin x \sin y)(\cos x \cos y+\sin x \sin y) \\ & = \cos^2 x \cos^2 y+\bcancel{\sin x \sin y \cos x \cos y}- \bcancel{\sin x \sin y \cos x \cos y} -\sin^2 x \sin^2 y \\ & = \cos^2 x \cos^2 y-\sin^2 x \sin^2 y \\ & = \cos^2 x(1-\sin^2 y)-(1-\cos^2 x) \sin^2 y \\ & = \cos^2 x-\cancel{\cos^2 x \sin^2 y}-\sin^2 y+\cancel{\cos^2 x \sin^2 y} \\ & = \cos^2 x-\sin^2 y \end{aligned}$$Therefore, it is proven that $\cos (x+y) \cos (x-y) = \cos^2 x-\sin^2 y$.

Use the angle relation and the following trigonometric identities.
$\begin{aligned} \cos x & = \sin (90^{\circ}-x) \\ \sin^2 x + \cos^2 x & = 1 \end{aligned}$
We can write
$$\begin{aligned} & \cos^2 1^{\circ} + \cos^2 2^{\circ} + \cdots + \cos^2 14^{\circ} + \cos^2 76^{\circ} + \cos^2 77^{\circ} + \cdots + \cos^2 90^{\circ} \\ & = \cos^2 1^{\circ} + \cos^2 2^{\circ} + \cdots + \cos^2 14^{\circ} + \sin^2 (90^{\circ}-76^{\circ}) + \sin^2 (90^{\circ}-77^{\circ}) + \cdots + \sin^2 (90^{\circ}-89^{\circ}) + \cos^2 90^{\circ} \\ & = (\cos^2 1^{\circ} +\sin^2 1^{\circ}) + (\cos^2 2^{\circ} + \sin^2 2^{\circ}) + \cdots + (\cos^2 14^{\circ} + \sin^2 14^{\circ}) + \color{red}{\cos^2 90^{\circ}} \\ & = \underbrace{1+1+\cdots+1}_{\text{there are 14}} + \color{red}{0} \\ & = 14 \end{aligned}$$Therefore, the result of the trigonometric expression is $\boxed{14}.$

Use the following logarithm property.
$\boxed{\log_a b + \log_a c = \log_a bc}$
Also use the complementary angle relation for tangent.
$\boxed{\tan x \tan (90^{\circ}-x) = \tan x \cot x = 1}$
We obtain
$$\begin{aligned} & \log_{10} (\tan 1^{\circ}) + \log_{10} (\tan 2^{\circ}) + \log_{10} (\tan 3^{\circ}) + \cdots + \log_{10} (\tan 88^{\circ}) + \log_{10} (\tan 89^{\circ}) \\ & = \log_{10} (\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \cdots \tan 88^{\circ} \cdot \tan 89^{\circ}) \\ & = \log_{10} ((\tan 1^{\circ} \tan 89^{\circ}) \cdot (\tan 2^{\circ} \tan 88^{\circ}) \cdots (\tan 44^{\circ} \tan 46^{\circ}) \cdot \tan 45^{\circ}) \\ & = \log_{10} (1 \cdot 1 \cdots 1 \cdot 1) \\ & = \log_{10} 1 = 0 \end{aligned}$$Therefore, the result of the calculation is $\boxed{0}.$

In the equation given above, multiply both sides by $\sin \left(\dfrac{x}{2^{2010}}\right)$, then use the double-angle identity $\sin x \cos x = \dfrac12 \sin 2x$ repeatedly up to $2010$ times.
$$\begin{aligned} \dfrac{1}{\sin \left(\dfrac{x}{2^{2010}}\right)} & = 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \cos \left(\dfrac{x}{2^{2010}}\right) \\ 1 & = 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \color{red}{\cos \left(\dfrac{x}{2^{2010}}\right) \sin \left(\dfrac{x}{2^{2010}}\right)} \\ 1 & = \dfrac12 \cdot 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \color{red}{\cos \left(\dfrac{x}{2^{2009}}\right) \sin \left(\dfrac{x}{2^{2009}}\right)} \\ 1 & = \left(\dfrac12\right)^2 \cdot 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \cos \color{red}{\left(\dfrac{x}{2^{2008}}\right) \sin \left(\dfrac{x}{2^{2008}}\right)} \\ \vdots & ~~~~~~~~~~~\vdots~~~~~~~~~~~~\vdots~~~~~~~~~~~~\vdots \\ 1 & = \left(\dfrac12\right)^{2009} \cdot 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \sin \left(\dfrac{x}{2}\right) \\ 1 & = \cancel{\left(\dfrac12\right)^{2010}} \cdot \cancel{2^{2010}} \sqrt2 \sin x \\ 1 & = \sqrt2 \sin x \\ \dfrac{1}{\sqrt2} & = \sin x \end{aligned}$$Since $0 \leq x \leq \pi$, we obtain $x = \dfrac{\pi}{4}$ or $x = \dfrac{3\pi}{4}.$