Tag: Special Angle

The length $OA = r$ is the hypotenuse length of a right triangle whose sides lie on the coordinate axes, so by using the Pythagorean Theorem, we obtain
$\begin{aligned} OA = r & = \sqrt{(-6)^2+8^2} \\ & = \sqrt{36+64} = \sqrt{100} = 10. \end{aligned}$
Since $\alpha$ lies in Quadrant II, the cosine value of angle alpha is negative, therefore
$\cos \alpha = -\dfrac{x}{r} = -\dfrac{6}{10} = -\dfrac35.$
Therefore, the value of $\boxed{\cos \alpha = -\dfrac35}.$
(Answer D)

For Quadrant III, the following angle relations apply:
$\begin{aligned} \sin (180^{\circ} + \alpha) & =-\sin \alpha \\ \cos (180^{\circ} + \alpha) & = -\cos \alpha \\ \tan (180^{\circ} + \alpha) & = \tan \alpha \end{aligned}$
Therefore, the trigonometric ratio equivalent to $\cos (180^{\circ} + \alpha)$ is $\boxed{-\cos \alpha}.$
(Answer D)

By substituting the values of the trigonometric ratios for special angles, we obtain
$\begin{aligned} \dfrac{\sin 60^{\circ}}{1 + \cos 60^{\circ}} & = \dfrac{\frac12\sqrt3}{1 + \frac12} \\ & = \dfrac{\frac{1}{\cancel{2}} \sqrt3}{\frac{3}{\cancel{2}}} \\ & = \dfrac{1}{3}\sqrt3 \end{aligned}$
The expression equivalent in value to $\dfrac13\sqrt3$ is $\boxed{\tan 30^{\circ}}.$
(Answer B)

Convert radians to degrees:
$\boxed{a~\text{rad} = a \times \dfrac{180^{\circ}} {\pi}}.$
Therefore,
$\begin{aligned} & \dfrac23\pi = \dfrac23\pi \times \dfrac{180^{\circ}} {\pi} = 120^{\circ} \\ & \dfrac73\pi = \dfrac73\pi \times \dfrac{180^{\circ}} {\pi} = 420^{\circ} \end{aligned}$
Thus,
$\begin{aligned} & \sin \dfrac23\pi + \sin \dfrac73 \pi \\ & = \sin 120^{\circ} + \sin 420^{\circ} \\ & = \sin (180-60)^{\circ} + \sin (360+60)^{\circ} \\ & = \sin 60^{\circ} + \sin 60^{\circ} \\ & = \dfrac12\sqrt3 +\dfrac12\sqrt3 = \sqrt3 \end{aligned}$
Therefore, the value of $\boxed{\sin \dfrac23\pi + \sin \dfrac73 \pi = \sqrt3}.$
(Answer E)

Using the concept of related-angle trigonometric ratios, we obtain
$\begin{aligned} & \dfrac{\sin 150^{\circ} + \sin 120^{\circ}}{\cos 210^{\circ}- \cos 300^{\circ}} \\ & = \dfrac{\sin (180-30)^{\circ} + \sin (180-60)^{\circ}}{\cos (180 + 30)^{\circ} -\cos (360 -60)^{\circ}} \\ & = \dfrac{\sin 30^{\circ} + \sin 60^{\circ}}{-\cos 30^{\circ} -\cos 60^{\circ}} \\ & = \dfrac{\dfrac12 + \dfrac12\sqrt3}{-\dfrac12\sqrt3- \dfrac12} \\ & = \dfrac{\cancel{\dfrac12 + \dfrac12\sqrt3}}{-\cancel{\left(\dfrac12\sqrt3 + \dfrac12\right)}} = -1 \end{aligned}$
Therefore, the value of $\boxed{\dfrac{\sin 150^{\circ} + \sin 120^{\circ}}{\cos 210^{\circ} – \cos 300^{\circ}} = -1}.$
(Answer E)

Using the concept of related-angle trigonometric ratios, we will calculate them one by one.
$$\begin{aligned} \cos 330^\circ & = \cos (-30^\circ) = cos 30^\circ = \dfrac12\sqrt3 \\ \tan (-315)^\circ & = \tan (-315 + 360)^\circ = \tan 45^\circ = 1 \\ \sin (-210)^\circ & = \sin (-210+360)^\circ = \sin 150^\circ = \dfrac12 \\ \cot 330^\circ & = \cot -30^\circ = -\cot 30^\circ = -\sqrt3 \end{aligned}$$Thus, we obtain
$$\begin{aligned} & \cos 330^{\circ} \tan (-315)^{\circ} -\sin (-210^{\circ}) \cot 330^{\circ} \\ & = \dfrac12\sqrt3 \cdot 1-\dfrac12 \cdot (-\sqrt3) \\ & = \dfrac12\sqrt3+\dfrac12\sqrt3 \\ & = \sqrt3. \end{aligned}$$Therefore, the value of $$\boxed{\cos 330^{\circ} \tan (-315)^{\circ} – \sin (-210^{\circ}) \cot 330^{\circ} = \sqrt3}.$$(Answer D)

Since $\alpha = \dfrac{3\pi}{4}$, then
$\begin{aligned} \sin \alpha & = \sin \dfrac{3\pi}{4} = \sin 135^{\circ} \\ & = \sin (180-45)^{\circ} \\ & = \sin 45^{\circ} = \dfrac12\sqrt2 \end{aligned}$
and
$\begin{aligned} \cos \alpha & = \cos \dfrac{3\pi}{4} = \cos 135^{\circ} \\ & = \cos (180-45)^{\circ} \\ & = -\cos 45^{\circ} = -\dfrac12\sqrt2 \end{aligned}$
Thus, the correct statement is
$$\boxed{\sin \alpha + \cos \alpha = \dfrac12\sqrt2 + \left(-\dfrac12\sqrt2\right) = 0}.$$(Answer C)

Notice that $\alpha$ lies in Quadrant III, so the tangent value is positive, while the sine value is negative.
Since $\tan \alpha = \dfrac34$, it can be assumed that the length of the opposite side is $3$, while the length of the adjacent side is $4$ (tan = op/adj) as shown in the following figure.

Thus, $\text{hyp} = \sqrt{3^2+4^2} = \sqrt{25}=5.$
Therefore, $\sin \alpha = -\dfrac{\text{op}} {\text{hyp}} = -\dfrac{3}{5}.$
(The sine value is negative when it lies in Quadrant III)
Therefore, the value of $\boxed{\sin \alpha =-\dfrac35}.$
(Answer C)

Notice that $\theta$ lies in Quadrant II, so the sine value is positive, while the cosine value is negative.
Since $\sin \theta = \dfrac56$, it can be assumed that the length of the opposite side is $5$, while the hypotenuse length is $6$ (sin = op/hyp) as shown in the following figure.
Thus, $\text{adj} = \sqrt{6^2-5^2} = \sqrt{36-25}=\sqrt{11}.$
Therefore,
$\cos \theta = -\dfrac{\text{adj}} {\text{hyp}} = -\dfrac{\sqrt{11}}{6} = -\frac16\sqrt{11}$
(The cosine value is negative when it lies in Quadrant II)
Therefore, the value of $\boxed{\cos \theta = -\frac16\sqrt{11}}.$
(Answer C)

Observe the following sketch.
Since the known side length is $BC$ (the opposite side) and the side length being asked is $AC$ (the hypotenuse), the trigonometric ratio used is sine (op/hyp).
$\begin{aligned} \sin 30^{\circ} & = \dfrac{BC}{AC} \\ \dfrac12 & = \dfrac{6}{AC} \\ AC & = 6 \times \dfrac21 = 12~\text{cm} \end{aligned}$
Therefore, the side length $\boxed{AC = 12~\text{cm}}.$
(Answer C)

By using the concept of related-angle trigonometric ratios, we obtain
$\begin{aligned} & \sin \left(\dfrac{\pi}{2} + 2x\right) + \sin \left(\dfrac{\pi}{2} – 2x\right) \\ & = \sin (90^{\circ} + 2x) + \sin (90^{\circ} – 2x) \\ & = \cos 2x + \cos 2x = 2 \cos 2x \end{aligned}$
(Answer B)

The equation $x+y+z=180^{\circ}$ is equivalent to $y+z=180^{\circ}-x$, therefore
$\begin{aligned} \sin \dfrac12(y+z) & = \sin \dfrac12(180^{\circ} -x) \\ & = \sin \left(90 -\dfrac12x\right) \\ & = \cos \dfrac12x \end{aligned}$
Therefore, the expression $\sin \dfrac12(y+z)$ is equivalent to $\boxed{\cos \dfrac12x}.$
(Answer E)

The sum of the three angles in a triangle is always $180^{\circ}$.
In triangle $ABC$, we have
$\begin{aligned} A + B + C & = 180^{\circ} \\ & \Leftrightarrow A + B & = 180^{\circ} -C. \end{aligned}$
Thus,
$\begin{aligned} \cot (A + B) & = \cot (180^{\circ} -C) \\ & = -\cot C. \end{aligned}$
Therefore, the value of $\boxed{\cot(A+B) = -\cot C}.$
(Answer E)

It is known that $\sin \alpha = \dfrac13\sqrt3 = \dfrac{\sqrt3}{3}.$
Since the sine of an angle is the ratio between the length of the opposite side and the hypotenuse in a right triangle, we may assume that they are respectively $\text{op} = \sqrt3$ and $\text{hyp} = 3$, so by using the Pythagorean Theorem, the length of the adjacent side is
$\begin{aligned} \text{adj} & = \sqrt{(\text{hyp})^2-(\text{op})^2} \\ & = \sqrt{(3)^2 – (\sqrt3)^2} \\ & = \sqrt{9-3} = \sqrt6. \end{aligned}$
Since $\alpha$ is an acute angle, all trigonometric ratios are positive.
Thus,
$$\begin{aligned} \tan \left(\dfrac{\pi}{2} – \alpha\right) + 3 \cos \alpha & = \tan (90^{\circ} -\alpha) + 3 \cos \alpha \\ & = \cot \alpha + 3 \cos \alpha \\ & = \dfrac{\text{adj}}{\text{op}} + 3 \times \dfrac{\text{adj}}{\text{hyp}} \\ & = \dfrac{\sqrt6}{\sqrt3} + \cancel{3} \times \dfrac{\sqrt6}{\cancel{3}} \\ & = \sqrt2 + \sqrt6 \\ & = \sqrt6 + \sqrt2. \end{aligned}$$Therefore, the value of $\boxed{\tan \left(\dfrac{\pi}{2} -\alpha\right) + 3 \cos \alpha = \sqrt6 + \sqrt2}.$
(Answer C)

It is known that $\tan \alpha = \dfrac12$.
Since the tangent of an angle is the ratio of the opposite side length to the adjacent side length in a right triangle, we may assume that they are respectively $\text{op} = 1$ and $\text{adj} = 3$ so that by using the Pythagorean Theorem, the hypotenuse length is
$\begin{aligned} \text{hyp} & = \sqrt{(\text{op})^2+(\text{adj})^2} \\ & = \sqrt{(1)^2 + (2)^2} \\ & = \sqrt{1+4} = \sqrt5. \end{aligned}$
Since $\alpha$ is an acute angle, all trigonometric ratios are positive.
Thus,
$$\begin{aligned} & 2 \sin \alpha -\sin \left(\alpha + \dfrac{\pi}{2}\right) + \cos (\pi -\alpha) \\ & = 2 \sin \alpha -\sin \left(\alpha + 90^{\circ}\right) + \cos (180^{\circ}- \alpha) \\ & = 2 \sin \alpha -\cos \alpha -\cos \alpha \\ & = 2 \sin \alpha -2 \cos \alpha \\ & = 2\left(\sin \alpha-\cos \alpha\right) \\ & = 2 \left(\dfrac{\text{op}}{\text{hyp}} -\dfrac{\text{adj}}{\text{hyp}}\right) \\ & = 2\left(\dfrac{1}{\sqrt5} -\dfrac{2}{\sqrt5}\right) = -\dfrac{2}{\sqrt5} = -\dfrac25\sqrt5. \end{aligned}$$Therefore, the value of $$\boxed{2 \sin \alpha -\sin \left(\alpha + \dfrac{\pi}{2}\right) + \cos (\pi -\alpha) = -\dfrac25\sqrt5}.$$(Answer D)

In right triangle $ABC$, we have
$\cos \theta = \dfrac{AB}{AC} = \dfrac{AB}{p} \Leftrightarrow AB = p \cos \theta.$
In right triangle $ABD$, we have
$\begin{aligned} \cos \theta & = \dfrac{AD}{AB} \\ AD & = AB \cos \theta \\ & = (p \cos \theta) \cos \theta = p \cos^2 \theta. \end{aligned}$
In right triangle $ADE$, we have
$\begin{aligned} \sin \theta & = \dfrac{DE}{AD} \\ DE & = AD \sin \theta \\ & = (p \cos^2 \theta) \sin \theta = p \sin \theta \cos^2 \theta. \end{aligned}$
Therefore, the length $\boxed{DE = p \sin \theta \cos^2 \theta}.$
(Answer C)

It is known that
$\begin{aligned} 1~\text{rad} & \approx 57,\!3^{\circ} \\ 2~\text{rad} & \approx 114,\!6^{\circ} \\ 3~\text{rad} & \approx 171,\!9^{\circ}. \end{aligned}$
This means,
$$\begin{aligned} \sin 1 & \approx \sin 57,\!3^{\circ} \\ \sin 2 & \approx \sin 114,\!6^{\circ} = \sin (180-65,\!4)^{\circ} = \sin 65,\!4^{\circ} \\ \sin 3 & \approx \sin 171,\!9^{\circ} = \sin (180-8,\!1)^{\circ} = \sin 8,\!1^{\circ}. \end{aligned}$$In Quadrant I, the larger the angle, the larger the sine value approaching $1$. Thus, the order of the values written as an inequality is $\sin 8,1^{\circ} < \sin 57,3^{\circ} < \sin 65,4^{\circ}$ or in radians written as $\boxed{\sin 3 < \sin 1 < \sin 2}.$
(Answer E)

Using the concept of angle relations in trigonometric ratios,
$\boxed{\begin{aligned} \cot x & = \tan (90^{\circ} -x) \\ \sin x & = \cos (90^{\circ} -x) \end{aligned}},$
we obtain
$$\begin{aligned} & \tan 25^{\circ} \cdot \tan 65^{\circ} -\dfrac{\sin 25^{\circ}}{\cos 65^{\circ}} \\ & = \tan 25^{\circ} \cdot \tan (90^{\circ}-25^{\circ})-\dfrac{\sin 25^{\circ}}{\cos (90^{\circ}-25^{\circ})} \\ & = \tan 25^{\circ} \cdot \cot 25^{\circ} -\dfrac{\sin 25^{\circ}}{\sin 25^{\circ}} \\ & = 1 -1 = 0 \end{aligned}$$Therefore, the value of $\boxed{\tan 25^{\circ} \cdot \tan 65^{\circ}- \dfrac{\sin 25^{\circ}}{\cos 65^{\circ}} = 0}.$
(Answer C)

Using the concept of angle relations in trigonometric ratios,
$\boxed{\begin{aligned} \sin (90^{\circ}- x) & = \cos x \\ \cot (90^{\circ} + x) & = -\tan x \\ \cos (180^{\circ} – x) & = -\cos x \\ \sin (180^{\circ}+x) & = -\sin x \end{aligned}},$
we obtain
$$\begin{aligned} & \cos 110^{\circ} \cdot \cot 160^{\circ} + \sin 200^{\circ} \\ & = \cos (180^{\circ}-70^{\circ}) \cdot \cot (90^{\circ} + 70^{\circ}) + \sin (180^{\circ}+20^{\circ}) \\ & = (-\cos 70^{\circ})(-\tan 70^{\circ})-\sin 20^{\circ} \\ & = \cancel{\cos 70^{\circ}} \cdot \dfrac{\sin 70^{\circ}}{\cancel{\cos 70^{\circ}}} -\sin (90^{\circ}-70^{\circ}) \\ & = \sin 70^{\circ}- \cos 70^{\circ} = p-q \end{aligned}$$Therefore, the value of $\boxed{\cos 110^{\circ} \cdot \cot 160^{\circ} + \sin 200^{\circ} = p-q}.$
(Answer A)

Answer a)
Since triangle $DEF$ is right-angled at $F$, the measure of angle $F = 90^{\circ}$ so that
$\begin{aligned} \cos (D + F) & = p \\ \cos (D + 90^{\circ}) & = p \\ -\sin D & = p \\ \sin D & = -p \end{aligned}$
Therefore, the value of $\boxed{\sin D = -p}.$
Answer b)
The sum of the angles in a triangle is $180^{\circ}$. In other words, it can be written as
$\begin{aligned} D + E +F & = 180^{\circ} \\ \Leftrightarrow E & = 180^{\circ} -D -F \end{aligned}$
Since the measure of angle $F = 90^{\circ}$, then $E = 90^{\circ} -D$ so that
$\cos E = \cos (90^{\circ} -D) = \sin D = -p.$
Therefore, the value of $\boxed{\cos E = -p}..$

In right triangle $ABD$, the length of $AD$ can be determined using the cosine trigonometric ratio.
$\begin{aligned} \cos 45^{\circ} & = \dfrac{AB}{BD} \\ \dfrac12\sqrt2 & = \dfrac{10}{BD} \\ BD & = 10 \times \dfrac{2}{\sqrt2} = 10\sqrt2~\text{cm} \end{aligned}$
In right triangle $BCD$, the length of $CD$ can also be determined using the cosine trigonometric ratio.
$\begin{aligned} \cos 60^{\circ} & = \dfrac{CD}{BD} \\ \dfrac12 & = \dfrac{CD}{10\sqrt2} \\ CD & = 10\sqrt2 \times \dfrac12 = 5\sqrt2~\text{cm} \end{aligned}$
Therefore, the length of $\boxed{CD = 5\sqrt2~\text{cm}}.$