Trigonometry is one of the branches of mathematics studied by high school students or students at the equivalent level. Trigonometry is a branch of mathematics that studies the relationship between side lengths and angle measures in a triangle. In trigonometry, there are six terms known as trigonometric ratios, namely sine, cosine, tangent, cosecant, secant, and cotangent. The first three trigonometric ratios (sine, cosine, and tangent) are the most commonly used.
Trigonometry has identities. A trigonometric identity is an open statement in the form of an equation involving trigonometric ratios that holds true for every chosen variable. One of the most well-known trigonometric identities is the Pythagorean Identity, namely $\sin^2 x + \cos^2 x = 1$. Trigonometric identities are derived from certain definitions or theorems, so their validity must be proven.
In such proofs, the phrase “proof from the left-hand side” is often used. This means that we begin the proof using the expression on the left-hand side of the equation in order to obtain the expression on the right-hand side. It is also possible to start from the right-hand side to derive the left-hand side. Usually, the proof starts with the more complicated expression to obtain the simpler one.
Therefore, below are several problems on proving trigonometric identities that can be used as learning references to gain a deeper understanding of the topic.
Today Quote
Problem Number 1
Prove the following trigonometric identity.
$$\dfrac{2 -\sec^2 A} {\sec^2 A} = 1 -2 \sin^2 A$$
Trigonometric identities used:
$$\boxed{\begin{aligned} \cos x & = \dfrac{1}{\sec x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$\begin{aligned} \dfrac{2 -\sec^2 A} {\sec^2 A} & = \dfrac{2}{\sec^2 A}- \dfrac{\sec^2 A} {\sec^2 A} \\ & = 2(\cos^2 A) -1 \\ & = 2(1 -\sin^2 A) -1 \\ & = 1 -2 \sin^2 A \end{aligned}$
Therefore, it is proven that $\dfrac{2 -\sec^2 A} {\sec^2 A} = 1 -2 \sin^2 A.$
Problem Number 2
Prove the following trigonometric identity.
$$(\sin A + \cos A)^2 -(\sin A -\cos A)^2 = 4 \sin A \cos A$$
Proof from the left-hand side:
$$\begin{aligned} & (\sin A + \cos A)^2 -(\sin A -\cos A)^2 \\ =& (\cancel{\sin^2 A} + 2 \sin A \cos A + \bcancel{\cos^2 A} ) \\ & -(\cancel{\sin^2 A} -2 \sin A \cos A + \bcancel{\cos^2 A}) \\ = & 2 \sin A \cos A -(-2 \sin A \cos A) \\ = & 4 \sin A \cos A \end{aligned}$$Therefore, it is proven that $$(\sin A + \cos A)^2 -(\sin A -\cos A)^2 = 4 \sin A \cos A.$$
Problem Number 3
Prove the following trigonometric identity.
$$\dfrac{\sec^2 \theta -1}{\sec^2 \theta} = \sin^2 \theta$$
Identities used:
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x &= 1 && (\text{Pythagorean Identity}) \\ \dfrac{1}{\sec x} & = \cos x && (\text{Reciprocal Identity}) \end{aligned}}$$Proof from the left-hand side:
$\begin{aligned} \dfrac{\sec^2 \theta -1}{\sec^2 \theta} & = \dfrac{\sec^2 \theta} {\sec^2 \theta} -\dfrac{1}{\sec^2 \theta} \\ & = 1 -\cos^2 \theta \\ & = \sin^2 \theta \end{aligned}$
Therefore, it is proven that $\dfrac{\sec^2 \theta -1}{\sec^2 \theta} = \sin^2 \theta.$
Problem Number 4
Prove the following trigonometric identity.
$$\dfrac{\sin A} {1 -\cos A} = \dfrac{1 + \cos A} {\sin A}$$
The identity used is the Pythagorean Identity, namely
$\boxed{\sin^2 x + \cos^2 x = 1}$
Proof from the left-hand side:
$\begin{aligned} \dfrac{\sin A} {1 -\cos A} & = \dfrac{\sin A} {1 -\cos A} \times \dfrac{1 + \cos A} {1 + \cos A} \\ & = \dfrac{\sin A(1 + \cos A)} {1 -\cos^2 A} \\ & = \dfrac{\cancel{\sin A}(1 + \cos A)} {\sin^{\cancel{2}} A} \\ & = \dfrac{1 + \cos A} {\sin A} \end{aligned}$
Therefore, it is proven that $\dfrac{\sin A} {1 -\cos A} = \dfrac{1 + \cos A} {\sin A}.$
Problem Number 5
Prove the following trigonometric identity.
$$\sin A + \cos A \cot A = \csc A$$
Identities used:
$$\boxed{\begin{aligned} \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \sin A + \cos A \cot A & = \sin A + \cos A \left(\dfrac{\cos A} {\sin A}\right) \\ & = \sin A + \dfrac{\cos^2 A} {\sin A} \\ & = \dfrac{\sin^2 A + \cos^2 A} {\sin A} \\ & = \dfrac{1}{\sin A} = \csc A \end{aligned}$$Therefore, it is proven that $\sin A + \cos A \cot A = \csc A.$
Problem Number 6
Prove the following trigonometric identity.
$$\csc A + \cot A = \dfrac{\sin A} {1- \cos A}$$
Identities used:
$$\boxed{\begin{aligned} \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the right-hand side:
$\begin{aligned} \dfrac{\sin A} {1- \cos A} & = \dfrac{\sin A} {1-\cos A} \times \dfrac{1 + \cos A} {1 + \cos A} \\ & = \dfrac{\sin A(1 + \cos A)} {1 -\cos^2 A} \\ & = \dfrac{\cancel{\sin A}(1 + \cos A)} {\sin^{\cancel{2}} A} \\ & = \dfrac{1 + \cos A} {\sin A} \\ & = \csc A + \cot A \end{aligned}$
Therefore, it is proven that $\csc A + \cot A = \dfrac{\sin A} {1- \cos A}.$
Problem Number 7
Prove the following trigonometric identity.
$$\sin A \csc A -\sin^2 A = \cos^2 A$$
Trigonometric identities used:
$$\boxed{\begin{aligned} \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cos^2 x + \sin^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \sin A \csc A -\sin^2 A & = \sin A \cdot \dfrac{1}{\sin A} -\sin^2 A \\ & = 1 -\sin^2 A = \cos^2 A \end{aligned}$$Therefore, it is proven that $\sin A \csc A -\sin^2 A = \cos^2 A.$
Problem Number 8
Prove the following trigonometric identity.
$$(\csc A + \cot A)(1 -\cos A) = \sin A$$
Trigonometric identities used:
$$\boxed{\begin{aligned} \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cot x & = \dfrac{\cos x}{\sin x} && (\text{Quotient Identity}) \\ \cos^2 x + \sin^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} & (\csc A + \cot A)(1 -\cos A) \\ & = \csc A -\csc A \cos A + \cot A -\cot A \cos A \\ & = \dfrac{1}{\sin A} -\dfrac{\cos A}{\sin A} + \cot A- \dfrac{\cos A}{\sin A} \cdot \cos A \\ & = \dfrac{1}{\sin A} – \cot A + \cot A- \dfrac{\cos^2 A}{\sin A} \\ & = \dfrac{1 -\cos^2 A}{\sin A} \\ & = \dfrac{\sin^2 A}{\sin A} = \sin A \end{aligned}$$Therefore, it is proven that $(\csc A + \cot A)(1 -\cos A) = \sin A.$
Problem Number 9
Prove the following trigonometric identity.
$$\tan^2 A -\sin^2 A = \tan^2 A \sin^2 A$$
Trigonometric identities used:
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x}{\cos x} && (\text{Quotient Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \tan^2 x & = \sec^2 x -1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \tan^2 A -\sin^2 A & = \dfrac{\sin^2 A}{\cos^2 A} -\sin^2 A \\ & = \sin^2 A \left(\dfrac{1}{\cos^2 A}-1\right) \\ & = \sin^2 A(\sec^2 A -1) \\ & = \sin^2 A \tan^2 A \\ & = \tan^2 A \sin^2 A \end{aligned}$$Therefore, it is proven that $\tan^2 A -\sin^2 A = \tan^2 A \sin^2 A.$
Problem Number 10
Prove the following trigonometric identity.
$$\tan A \cos^4 A + \cot A \sin^4 A = \sin A \cos A$$
Identities used:
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \\ \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$\begin{aligned} & \tan A \cos^4 A + \cot A \sin^4 A \\ = & \dfrac{\sin A} {\cos A} \cdot \cos^4 A + \dfrac{\cos A} {\sin A} \cdot \sin^4 A \\ = & \sin A \cos^3 A + \cos A \sin^3 A \\ = & \sin A \cos A(\cos^2 A + \sin^2 A) \\ = & \sin A \cos A(1) = \sin A \cos A \end{aligned}$
Therefore, it is proven that $\tan A \cos^4 A + \cot A \sin^4 A$ $= \sin A \cos A.$
Problem Number 11
Prove the following trigonometric identity.
$$\dfrac{\sin^2 A} {\cos^2 A} -\dfrac{\cos^2 A} {\sin^2 A} = \sec^2 A -\csc^2 A$$
Trigonometric identities used:
$$\boxed{\begin{aligned} \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} & \dfrac{\sin^2 A} {\cos^2 A} -\dfrac{\cos^2 A} {\sin^2 A} \\ & = \dfrac{\sin^4 A -\cos^4 A} {\cos^2 A \sin^2 A} \\ & = \dfrac{(\sin^2 A -\cos^2 A)(\sin^2 A + \cos^2 A)} {\cos^2 A \sin^2 A} \\ & = \dfrac{(\sin^2 A -\cos^2 A)(1)} {\cos^2 A~\sin^2 A} \\ & = \dfrac{\cancel{\sin^2 A}} {\cos^2 A~\cancel{\sin^2 A}} -\dfrac{\bcancel{\cos^2 A}} {\bcancel{\cos^2 A} \sin^2 A} \\ & = \dfrac{1}{\cos^2 A} -\dfrac{1}{\sin^2 A} \\ & = \sec^2 A -\csc^2 A \end{aligned}$$Therefore, it is proven that $\dfrac{\sin^2 A} {\cos^2 A} -\dfrac{\cos^2 A} {\sin^2 A} = \sec^2 A -\csc^2 A.$
Problem Number 12
Prove the following trigonometric identity.
$\dfrac{\sin^2 A -\sin^2 B} {\cos^2 A \cos^2 B} = \tan^2 A -\tan^2 B$
Trigonometric identities used:
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \sec^2 x & = 1 + \tan^2 x && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} & \dfrac{\sin^2 A -\sin^2 B} {\cos^2 A \cos^2 B} \\ & = \dfrac{\sin^2 A}{\cos^2 A \cos^2 B} -\dfrac{\sin^2 B} {\cos^2 A \cos^2 B}\\ & = \dfrac{\sin^2 A} {\cos^2 A} \cdot \dfrac{1}{\cos^2 B} -\dfrac{\sin^2 B} {\cos^2 B} \cdot \dfrac{1}{\cos^2 A} \\ & = \tan^2 A(\sec^2 B) -\tan^2 B(\sec^2 A) \\ & = \tan^2 A(1 + \tan^2 B) -\tan^2 B(1 + \tan^2 A) \\ & = \tan^2 A + \cancel{\tan^2 A \tan^2 B} -\tan^2 B -\cancel{\tan^2 A \tan^2 B} \\ & = \tan^2 A- \tan^2 B \end{aligned}$$Therefore, it is proven that $\dfrac{\sin^2 A -\sin^2 B} {\cos^2 A \cos^2 B} = \tan^2 A- \tan^2 B.$
Problem Number 13
Prove the following trigonometric identity.
$\cos^2 A + \cot^2 A \cos^2 A = \cot^2 A$
Trigonometric identities used:
$$\boxed{\begin{aligned} 1 + \cot^2 x & = \csc^2 x && (\text{Pythagorean Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \end{aligned}}$$Proof from the left-hand side:
$\begin{aligned} & \cos^2 A + \cot^2 A \cos^2 A \\ & = \cos^2 A(1 + \cot^2 A) \\ & = \cos^2 A(\csc^2 A) \\ & = \cos^2 A \cdot \dfrac{1}{\sin^2 A} \\ & = \cot^2 A \end{aligned}$
Therefore, it is proven that $\cos^2 A + \cot^2 A \cos^2 A = \cot^2 A.$
Problem Number 14
Prove the following trigonometric identity
$\dfrac{1 + \sec A} {\tan A + \sin A} = \csc A$
Trigonometric identities used:
$$\boxed{\begin{aligned} \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$Proof from the left-hand side:
$\begin{aligned} \dfrac{1 + \sec A} {\tan A + \sin A} & = \dfrac{1 + \frac{1}{\cos A}} {\frac{\sin A} {\cos A} + \sin A} \\ & = \dfrac{\cos A + 1}{\sin A + \sin A \cos A} \\ & = \dfrac{\cancel{\cos A + 1}} {\sin A(\cancel{1 + \cos A}) } \\ & = \dfrac{1}{\sin A} = \csc A \end{aligned}$
Therefore, it is proven that $\dfrac{1 + \sec A} {\tan A + \sin A} = \csc A.$
Problem Number 15
Prove the following trigonometric identity.
$$\dfrac{1 + \sin x} {1 -\sin x} = (\sec x + \tan x)^2$$
Trigonometric identities used:
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \dfrac{1 + \sin x} {1 -\sin x} & = \dfrac{1 + \sin x} {1 -\sin x} \times \dfrac{1 + \sin x} {1 + \sin x} \\ & = \dfrac{1 + 2 \sin x + \sin^2 x} {1- \sin^2 x} \\ & = \dfrac{1 + 2 \sin x + \sin^2 x} {\cos ^2 x} \\ & = \dfrac{1}{\cos^2 x} + \dfrac{2 \sin x} {\cos^2 x} + \dfrac{\sin^2 x} {\cos^2 x} \\ & = \sec^2 x + 2 \sec x \tan x + \tan^2 x \\ & = (\sec x + \tan x)^2 \end{aligned}$$Therefore, it is proven that $\dfrac{1 + \sin x} {1 -\sin x} = (\sec x + \tan x)^2.$
Problem Number 16
Prove the following trigonometric identity.
$$\tan A = \dfrac12 \sin 2A(1+\tan^2 A)$$
Use the following trigonometric identities.
$\boxed{\begin{aligned} \sin 2x & = 2 \sin x \cos x \\ \tan^2 x + 1 & = \sec^2 x \end{aligned} }$
Proof from the right-hand side:
$\begin{aligned} & \dfrac12 \sin 2A(1+\tan^2 A) \\ & = \dfrac12(2 \sin A \cos A) (1+\tan^2 A) \\ & = (\sin A \cos A) (1+\tan^2 A) \\ & = (\sin A \cos A) (\sec^2 A) \\ & = \dfrac{\sin A \cancel{\cos A}} {\cancelto{\cos A} {\cos^2 A}} \\ & = \dfrac{\sin A} {\cos A} = \tan A \end{aligned}$
Therefore, it is proven that $\tan A = \dfrac12 \sin 2A(1+\tan^2 A).$
Problem Number 17
Prove the following trigonometric identity.
$$\tan^2 A = 1 -\cos 2A(1 + \tan^2 A)$$
Use the following trigonometric identities.
$\boxed{\begin{aligned} 1+\tan^2 x & = \sec^2 x \\ \cos 2x & = 1 -2 \sin^2 x \\ \sec x & = \dfrac{1}{\cos x} \end{aligned}}$
Proof from the right-hand side:
$\begin{aligned} & 1 – \cos 2A(1 + \tan^2 A) \\ & = 1 -\cos 2A(\sec^2 A) \\ & = 1- \dfrac{1 -2 \sin^2 A}{\cos^2 A} \\ & = 1 -\dfrac{1}{\cos^2 A} + \dfrac{2 \sin^2 A} {\cos^2 A} \\ & = 1 -\sec^2 A + 2 \tan^2 A \\ & = 1 -(1+\tan^2 A) + 2 \tan^2 A \\ & = \tan^2 A \end{aligned}$
Therefore, it is proven that $\tan^2 A = 1 -\cos 2A(1 + \tan^2 A).$
Problem Number 18
Prove the following trigonometric identity.
$$\tan^6 A = \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \sec^2 A -1$$
Use the following trigonometric identity.
$\boxed{\tan^2 x + 1 = \sec^2 x}$
Proof from the right-hand side:
$$\begin{aligned} & \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \sec^2 A -1 \\ & = \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \tan^2 A \\ & = \tan^2 A(\tan^2 A \sec^2 A -\sec^2 A + 1) \\ & = \tan^2 A((\sec^2 A)(\tan^2 A- 1)+1) \\ & = \tan^2 A((\tan^2 A + 1)(\tan^2 A -1)+1) \\ & = \tan^2 A(\tan^4 A-1 + 1) \\ & = \tan^6 A \end{aligned}$$Therefore, it is proven that $$\tan^6 A = \tan^4 A \cdot \sec^2 A -\tan^2 A \cdot \sec^2 A + \sec^2 A -1.$$
Problem Number 19
Prove the following trigonometric identity.
$$\tan (A -B) = \dfrac{\sin 2A -\sin 2B} {\cos 2A + \cos 2B}$$
Use the following trigonometric identities.
$$\boxed{\begin{aligned} \sin x -\sin y & = 2 \cos \dfrac12(x + y) \sin \dfrac12(x-y) \\ \cos x + \cos y & = 2 \cos \dfrac12(x + y) \cos \dfrac12(x-y) \\ \tan x & = \dfrac{\sin x} {\cos x} \end{aligned}}$$Proof from the right-hand side:
$$\begin{aligned} \dfrac{\sin 2A -\sin 2B} {\cos 2A + \cos 2B} & = \dfrac{2 \cos \dfrac12(2A + 2B) \sin \dfrac12(2A-2B)} {2 \cos \dfrac12(2A+2B) \cos \dfrac12(2A-2B)} \\ & = \dfrac{\cancel{2 \cos (A+B)} \sin (A-B)} {\cancel{2 \cos (A+B)} \cos (A-B)} \\ & = \dfrac{\sin (A-B)} {\cos (A-B)} = \tan (A-B) \end{aligned}$$Therefore, it is proven that $\tan (A -B) = \dfrac{\sin 2A -\sin 2B} {\cos 2A + \cos 2B}.$
Problem Number 20
Prove the following trigonometric identity.
$$\cot 2A = \dfrac{\sin A -\sin 3A} {\cos 3A -\cos A}$$
Use the following trigonometric identities.
$$\boxed{\begin{aligned} \sin x -\sin y & = 2 \cos \dfrac12(x + y) \sin \dfrac12(x-y) \\ \cos x -\cos y & = -2 \sin \dfrac12(x + y) \sin \dfrac12(x-y) \\ \sin (-x) & = -\sin x \\ \cot x & = \dfrac{\cos x} {\sin x} \end{aligned}}$$Proof from the right-hand side:
$$\begin{aligned} \dfrac{\sin A -\sin 3A} {\cos 3A -\cos A} & = \dfrac{2 \cos \dfrac12(A + 3A) \sin \dfrac12(A-3A)} {-2 \sin \dfrac12(3A + A) \sin \dfrac12(3A- A)} \\ & = \dfrac{2 \cos 2A \sin (-A)} {-2 \sin 2A \sin A} \\ & = \dfrac{\cancel{-2 \sin A} \cos 2A} {\cancel{-2 \sin A} \sin 2A} \\ & = \dfrac{\cos 2A} {\sin 2A} = \cot 2A \end{aligned}$$Therefore, it is proven that $\cot 2A = \dfrac{\sin A -\sin 3A} {\cos 3A -\cos A}.$
Problem Number 21
Prove the following trigonometric identity.
$$\dfrac{1 -\cos 2A + \sin A} {\sin 2A + \cos A} =\tan A$$
The trigonometric identities used are:
$$\boxed{\begin{aligned} \cos 2x & = \cos^2 x – \sin^2 x && (\text{Double-Angle Identity}) \\ \sin 2x &= 2 \sin x \cos x && (\text{Double-Angle Identity}) \\ \sin^2 x + \cos^2 x &= 1 && (\text{Pythagorean Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \dfrac{1 -\cos 2A + \sin A} {\sin 2A + \cos A} & = \dfrac{1- (\cos^2 A -\sin^2 A) + \sin A} {(2 \sin A \cos A) + \cos A} \\ & = \dfrac{(1- \cos^2 A) + \sin^2 A + \sin A} {\cos A(2 \sin A + 1)} \\ & = \dfrac{\sin^2 A + \sin^2 A + \sin A} {\cos A(2 \sin A + 1)} \\ & = \dfrac{\sin A\cancel{(2 \sin A + 1)}} {\cos A\cancel{(2 \sin A + 1)}} \\ & = \dfrac{\sin A} {\cos A} = \tan A \end{aligned}$$Therefore, it is proven that $\dfrac{1 -\cos 2A + \sin A} {\sin 2A + \cos A} =\tan A.$
Problem Number 22
Prove the following trigonometric identity.
$$\csc 2A + \cot 2A = \cot A$$
The trigonometric identities used are:
$$\boxed{\begin{aligned} \cos 2x &= \cos^2 x -\sin^2 x && (\text{Double-Angle Identity}) \\ \sin 2x & = 2 \sin x \cos x && (\text{Double-Angle Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \csc x & = \dfrac{1}{\sin x} && (\text{Reciprocal Identity}) \\ \cot x & = \dfrac{\cos x} {\sin x} && (\text{Quotient Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \csc 2A + \cot 2A & = \dfrac{1}{\sin 2A} + \dfrac{\cos 2A} {\sin 2A} \\ & = \dfrac{1 + (\cos^2 A -\sin^2 A)} {2 \sin A \cos A} \\ & = \dfrac{(1- \sin^2 A) + \cos^2 A} {2 \sin A \cos A} \\ & = \dfrac{2 \cos^2 A} {2 \sin A \cos A} \\ & = \dfrac{\cos A} {\sin A} = \cot A \end{aligned}$$Therefore, it is proven that $\csc 2A + \cot 2A = \cot A.$
Problem Number 23
Prove the following trigonometric identity.
$$\dfrac{2 \sin (A-B)} {\cos (A-B)-\cos (A+B)} = \cot B -\cot A$$
Recall the sum and difference formulas:
$\boxed{\begin{aligned} \sin (x \pm y) & = \sin x \cos y \pm \cos x \sin y \\ \cos (x \pm y) & = \cos x \cos y \mp \sin x \sin y \end{aligned}}$
The trigonometric identity used is:
$ \cot x = \dfrac{\cos x} {\sin x} $
Proof from the left-hand side:
$$\begin{aligned} & \dfrac{2 \sin (A-B)} {\cos (A-B)-\cos (A+B)} \\ & = \dfrac{2 (\sin A \cos B -\sin B \cos A)} {(\cancel{\cos A \cos B} + \sin A \sin B)- (\cancel{\cos A \cos B} -\sin A \sin B)} \\ & = \dfrac{\cancel{2}(\sin A \cos B -\sin B \cos A)} {\cancel{2} \sin A \sin B} \\ & = \dfrac{\sin A \cos B -\sin B \cos A} {\sin A \sin B} \\ & = \dfrac{\cancel{\sin A} \cos B} {\cancel{\sin A} \sin B}- \dfrac{\cancel{\sin B} \cos A} {\sin A \cancel{\sin B}} \\ & = \dfrac{\cos B} {\sin B} -\dfrac{\cos A} {\sin A} \\ & = \cot B -\cot A \end{aligned}$$Therefore, it is proven that $$\dfrac{2 \sin (A-B)} {\cos (A-B)-\cos (A+B)} = \cot B -\cot A.$$
Problem Number 24
Prove the following trigonometric identity.
$$\tan A -\cot A = \dfrac{1 -2 \cos^2 A} {\sin A \cos A}$$
The trigonometric identities used are:
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \\ \cot x & = \dfrac{\cos x}{\sin x} && (\text{Quotient Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \tan A -\cot A & = \dfrac{\sin A} {\cos A} -\dfrac{\cos A} {\sin A} \\ & = \dfrac{\sin^2 A -\cos^2 A} {\sin A \cos A} \\ & = \dfrac{(1 -\cos^2 A)- \cos^2 A} {\sin A \cos A} \\ & = \dfrac{1 -2 \cos^2 A} {\sin A \cos A} \end{aligned}$$Therefore, it is proven that $\tan A -\cot A = \dfrac{1 -2 \cos^2 A} {\sin A \cos A}.$
Problem Number 25
Prove the following trigonometric identity.
$$1 -\tan^2 A= \cos 2A \sec^2 A$$
The trigonometric identities used are:
$$\boxed{\begin{aligned} \cos 2x & = \cos^2 x -\sin^2 x && (\text{Double-Angle Identity}) \\ \sec x & =\dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$Proof from the right-hand side:
$$\begin{aligned} \cos 2A \sec^2 A & = (\cos^2 A -\sin^2 A) \cdot \dfrac{1}{\cos^2 A} \\ & = 1 -\dfrac{\sin^2 A} {\cos^2 A} \\ & = 1 -\tan^2 A \end{aligned}$$Therefore, it is proven that $1 -\tan^2 A= \cos 2A \sec^2 A.$
Problem Number 26
Prove the following trigonometric identity.
$$\dfrac{\sin \left(A + \frac{\pi} {4}\right)} {\cos \left(A + \frac{\pi} {4}\right)} + \dfrac{\cos \left(A + \frac{\pi} {4}\right)} {\sin \left(A + \frac{\pi} {4}\right)} = 2 \sec 2A$$
Recall the values of trigonometric ratios for angles in various quadrants:
$\sin (2x + 90^{\circ}) = \cos 2x.$
The trigonometric identities used are:
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \sin 2x & = 2 \sin x \cos x && (\text{Double-Angle Identity}) \\ \sec x & = \dfrac{1}{\cos x} && (\text{Reciprocal Identity}) \end{aligned}}$$Proof from the left-hand side:
$\begin{aligned} & \dfrac{\sin \left(A + \frac{\pi} {4}\right)} {\cos \left(A + \frac{\pi} {4}\right)} + \dfrac{\cos \left(A + \frac{\pi} {4}\right)} {\sin \left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{\sin^2 \left(A + \frac{\pi} {4}\right) + \cos^2 \left(A + \frac{\pi} {4}\right)}{\sin \left(A + \frac{\pi} {4}\right) \cdot \cos \left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{1}{\sin \left(A + \frac{\pi} {4}\right) \cdot \cos \left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{2}{\sin 2\left(A + \frac{\pi} {4}\right)} \\ & = \dfrac{2}{\sin \left(2A + \frac{\pi} {2}\right)} \\ & = \dfrac{2}{\cos 2A} = 2 \sec 2A \end{aligned}$
Therefore, it is proven that $$\dfrac{\sin \left(A + \frac{\pi} {4}\right)} {\cos \left(A + \frac{\pi} {4}\right)} + \dfrac{\cos \left(A + \frac{\pi} {4}\right)} {\sin \left(A + \frac{\pi} {4}\right)} = 2 \sec 2A.$$
Read Also: Problems and Solutions – Trigonometric Equations of the Form a cos x + b sin x = c
Problem Number 27
Prove the following trigonometric identity.
$$\dfrac{\sin 4A} {1 + \cos 4A} = \tan 2A$$
The trigonometric identities used are:
$$\boxed{\begin{aligned} \sin 2x & = 2 \sin x \cos x && (\text{Double-Angle Identity}) \\ \cos 2x & = \cos^2 x -\sin^2 x && (\text{Double-Angle Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \tan x & = \dfrac{\sin x} {\cos x} && (\text{Quotient Identity}) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} \dfrac{\sin 4A} {1 + \cos 4A} & = \dfrac{2 \sin 2A \cos 2A} {1 + (\cos^2 2A -\sin^2 2A)} \\ & = \dfrac{2 \sin 2A \cos 2A} {(1 -\sin^2 2A) + \cos^2 2A} \\ & = \dfrac{2 \sin 2A \cos 2A} {2 \cos^2 2A} \\ & = \dfrac{\sin 2A} {\cos 2A} = \tan 2A \end{aligned}$$Therefore, it is proven that $\dfrac{\sin 4A} {1 + \cos 4A} = \tan 2A.$
Problem Number 28
Prove the following trigonometric identity.
$$\begin{aligned} (\cos 2a + \cos 4a & + \cos 6a) \sin a \\ & = \cos 4a \sin 3a \end{aligned}$$
The trigonometric identities used are
$$\boxed{\begin{aligned} \cos A + \cos B & = 2 \cos \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A-B}{2}\right) \\ \cos A \sin B & = \dfrac12\left(\sin (A + B) + \sin (A-B)\right) \end{aligned}}$$Proof from the left-hand side:
$$\begin{aligned} & (\cos 2a + \cos 4a + \cos 6a) \sin a \\ & = \left(\cos 4a + 2 \cos \left(\dfrac{2a+6a}{2}\right) \cos \left(\dfrac{6a-2a}{2}\right)\right) \sin a \\ & = (\cos 4a + 2 \cos 4a \cos 2a) \sin a \\ & = \cos 4a(1 + 2 \cos 2a) \sin a \\ & = \cos 4a(\sin a + 2 \cos 2a \sin a) \\ & = \cos 4a(\sin a + \cancel{2} \cdot \dfrac{1}{\cancel{2}}(\sin (a+2a) + \sin (a-2a))) \\ & = \cos 4a(\cancel{\sin a} \sin 3a~ \cancel{\sin (-a)}) \\ & = \cos 4a \sin 3a \end{aligned}$$Therefore, it is proven that
$$(\cos 2a + \cos 4a + \cos 6a) \sin a = \cos 4a \sin 3a.$$
Problem Number 29
Prove the following trigonometric identity.
$$2 \sin A \cos (A + 30^{\circ}) = \sin(2A + 30^{\circ}) -\dfrac12$$
Use the following trigonometric identities.
$$\boxed{\begin{aligned} \cos (A + B) & = \cos A \cos B – \sin A \sin B \\ \sin 2A & = 2 \sin A \cos A \\ 1 -\cos 2A & = 2 \sin^2 A \\ \sin(A + B) & = \sin A \cos B + \sin B \cos A \end{aligned}}$$Proof of the left-hand side:
$$\begin{aligned} & 2 \sin A \cos (A + 30^{\circ}) \\ & = 2 \sin A(\cos A \cos 30^{\circ} -\sin A \sin 30^{\circ}) \\ & = 2 \sin A\left(\dfrac12\sqrt3 \cos A- \dfrac12 \sin A\right) \\ & = \dfrac12\sqrt3(2 \sin A \cos A) -\dfrac12(\sin^2 A) \\ & = \dfrac12\sqrt3(\sin 2A) -\dfrac12(1 -\cos 2A) \\ & = \dfrac12\sqrt3(\sin 2A)- \dfrac12 + \dfrac12 \cos 2A \\ & = (\sin 2A \cos 30^{\circ} + \cos 2A \sin 30^{\circ}) -\dfrac12 \\ & = \sin (2A + 30^{\circ}) -\dfrac12 \end{aligned}$$Thus, it is proven that
$$2 \sin A \cos (A + 30^{\circ}) = \sin(2A + 30^{\circ}) -\dfrac12.$$
Problem Number 30
Prove the following trigonometric identity.
$$2 \cos \left(\dfrac{\pi}{4} + A\right) \cos \left(\dfrac{3\pi}{4} -A \right) = -1 +\sin 2A$$
Use the following trigonometric identities.
$$\boxed{\begin{aligned} 2 \cos A \cos B & = \cos (A+B) + \cos (A-B) \\ \sin x & = \cos (90^{\circ} -x) = \cos (x -90^{\circ}) \end{aligned}}$$Proof of the left-hand side:
$\begin{aligned} & 2 \cos \left(\dfrac{\pi}{4} + A\right) \cos \left(\dfrac{3\pi}{4}- A \right) \\ & = \cos \left(\left(\dfrac{\pi}{4}+A\right) + \left(\dfrac{3\pi}{4}- A\right)\right) \\ & + \cos \left(\left(\dfrac{\pi}{4} + A\right) -\left(\dfrac{3\pi}{4} – A\right)\right) \\ & = \cos \pi + \cos \left(-\dfrac{\pi}{2} + 2A\right) \\ & = -1 + \sin 2A \end{aligned}$
Thus, it is proven that $$2 \cos \left(\dfrac{\pi}{4} + A\right) \cos \left(\dfrac{3\pi}{4} -A \right) = -1 +\sin 2A.$$
Problem Number 31
Prove the following trigonometric identity.
$\begin{aligned} \sin^4 x -\cos^4 x & = 1 -2(\sin x \cos x)^2\\ & -2 \cos^4 x \end{aligned}$
Use the following Pythagorean identity.
$\boxed{\sin^2 x + \cos^2 x = 1}$
Proof of the right-hand side:
$$\begin{aligned} & 1 -2(\sin x \cos x)^2 – 2 \cos^4 x \\ & = 1 -2 \sin^2 x \cos^2 x -2 \cos^4 x \\ & = (\sin^2 x + \cos^2 x) -2(1- \cos^2 x)(\cos^2 x) -2 \cos^4 x \\ & = (\sin^2 x + \cos^2 x) -2 \cos^2 x + \cancel{2 \cos^4 x -2 \cos^4 x} \\ & = \sin^2 x -\cos^2 x \\ & = (\sin^2 x -\cos^2 x)(1) \\ & = (\sin^2 x -\cos^2 x) (\sin^2 x + \cos^2 x) \\ & = \sin^4 x -\cos^4 x \end{aligned}$$Thus, it is proven that $$\sin^4 x -\cos^4 x = 1 -2(\sin x \cos x)^2-2 \cos^4 x.$$
Problem Number 32
Prove the following trigonometric identity.
$$\cos^6 x + 3 \sin^2 x \cos^2 x + \sin^6 x = 1$$
Use the result of the cubic binomial expansion and the following trigonometric identity.
$$\boxed{\begin{aligned} (a+b)^3 & = a^3 + 3a^2b + 3ab^2 + b^3 \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pyt}\text{hagorean Identity}) \end{aligned}}$$Proof of the right-hand side:
$$\begin{aligned} 1 & = 1^3 \\ & = (\sin^2 x + \cos^2 x)^3 \\ & = \sin^6 x + \cos^6 x + 3 \sin^4 x \cos^2 x + 3 \sin^2 x \cos^4 x \\ & = \sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x(\sin^2 x + \cos^2 x) \\ & = \sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x(1) \\ & = \cos^6 x + 3 \sin^2 x \cos^2 x + \sin^6 x \end{aligned}$$Thus, it is proven that $\cos^6 x + 3 \sin^2 x \cos^2 x + \sin^6 x = 1.$
Problem Number 33
Prove the following trigonometric identity.
$$\dfrac{\tan x-\sin x}{\sin^3 x} = \dfrac{1}{\cos x(1 + \cos x)}$$
Use the following trigonometric identities.
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x}{\cos x} && (\text{Ratio Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pyt}\text{hagorean Identity}) \end{aligned}}$$Proof of the left-hand side:
$$\begin{aligned} \dfrac{\tan x-\sin x}{\sin^3 x} & = \dfrac{\dfrac{\sin x}{\cos x}- \sin x}{\sin^3 x} \times \color{red}{\dfrac{\cos x}{\cos x}} \\ & = \dfrac{\sin x -\sin x \cos x}{\sin^3 x \cos x} \\ & = \dfrac{\cancel{\sin x}(1 -\cos x)}{\cancelto{\sin^2 x}{\sin^3 x} \cos x} \\ & = \dfrac{1-\cos x}{\sin^2 x \cos x} \\ & = \dfrac{1- \cos x}{(1 -\cos^2 x) \cos x} \\ & = \dfrac{\cancel{1- \cos x}}{\cancel{(1 -\cos x)}(1 + \cos x)\cos x} \\ & = \dfrac{1}{\cos x(1 + \cos x)} \end{aligned}$$Thus, it is proven that $\dfrac{\tan x- \sin x}{\sin^3 x} = \dfrac{1}{\cos x(1 + \cos x)}.$
Problem Number 34
Prove the following trigonometric identity.
$$\begin{aligned} & \dfrac{\cot x-\tan x}{(\cos x-\sin x)(\cot x+\tan x)} \\ & = \sin x + \cos x \end{aligned}$$
Use the following trigonometric identities.
$$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x}{\cos x} && (\text{Ratio Identity}) \\ \cot x & = \dfrac{\cos x}{\sin x} && (\text{Ratio Identity}) \\ \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \end{aligned}}$$Proof from the left-hand side:
$\begin{aligned} & \dfrac{\cot x-\tan x}{(\cos x-\sin x)(\cot x+\tan x)} \\ & = \dfrac{\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}}{(\cos x-\sin x)\left(\dfrac{\cos x}{\sin x}+ \dfrac{\sin x}{\cos x}\right)} \\ & = \dfrac{\dfrac{cos^2 x- \sin^2 x}{\cancel{\sin x \cos x}}}{(\cos x-\sin x)\left(\dfrac{\cos^2 x + \sin^2 x}{\cancel{\sin x \cos x}}\right)} \\ & = \dfrac{\cos^2 x- \sin^2 x}{(\cos x-\sin x)(\sin^2 x + \cos^2 x)} \\ & = \dfrac{(\cos x + \sin x)\cancel{(\cos x- \sin x)}}{\cancel{(\cos x-\sin x)}(1)} \\ & = \sin x + \cos x \end{aligned}$
Therefore, it is proven that $$\dfrac{\cot x-\tan x}{(\cos x-\sin x)(\cot x+\tan x)} = \sin x + \cos x.$$
Problem Number 35
Prove the following trigonometric identity.
$$\dfrac{(\tan x-1)(\tan x + \cot x)}{\tan x- \cot x} = \dfrac{\tan^2 x+1}{\tan x +1}$$
Use the following trigonometric identity.
$$\boxed{\tan x \cot x = 1~~~~~(\text{Reciprocal Identity})}$$Proof from the right-hand side:
$$\begin{aligned} \dfrac{\tan^2 x +1}{\tan x +1} & = \dfrac{\tan^2 x + \tan x \cot x}{\tan x + \tan x \cot x} \\ & = \dfrac{\cancel{\tan x}(\tan x + \cot x)}{\cancel{\tan x}(1 + \cot x)} \\ & = \dfrac{\tan x + \cot x}{1+\cot x} \times \color{red}{\dfrac{\tan x-\cot x}{\tan x-\cot x}} \\ & = \dfrac{(\tan x + \cot x)(\tan x -\cot x)}{(\tan x-\cot x)(1+\cot x)} \times \color{blue}{\dfrac{\tan x- 1}{\tan x-1}} \\ & = \dfrac{(\tan x + \cot x)(\tan x-\cot x)(\tan x-1)}{(\tan x-\cot x)(\tan x \cancel{-1 + \tan x \cos x} -\cot x)} \\ & = \dfrac{(\tan x + \cot x)\cancel{(\tan x-\cot x)}(\tan x-1)}{(\tan x-\cot x)\cancel{(\tan x – \cot x)}} \\ & = \dfrac{(\tan x -1)(\tan x + \cot x)}{\tan x-\cot x} \end{aligned}$$Therefore, it is proven that $$\dfrac{(\tan x-1)(\tan x + \cot x)}{\tan x- \cot x} = \dfrac{\tan^2 x+1}{\tan x +1}.$$
Problem Number 36
Prove that $\sin^6 A + \cos^6 A + \dfrac34 \sin^2 2A = 1$.
For cubic powers, the following expansion holds.
$$\begin{aligned} (a+b)^3 & = a^3+b^3+3a^2b+3ab^2 \\ \Rightarrow (a^2 + b^2)^3 & = a^6 + b^6+3a^4b^2+3a^2b^4 \\ & =a^6+b^6+3a^2b^2(a^2+b^2) \end{aligned}$$Use this fact together with the following trigonometric identities to prove the statement above.
$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 \\ \sin 2x & = 2 \sin x \cos x \end{aligned}}$
Thus, from the left-hand side we obtain
$$\begin{aligned} &\sin^6 A + \cos^6 A + \dfrac34 \sin^2 2A \\ & = (\sin^2 A + \cos^2 A)^3-3 \sin^2 A \cos^2 A(\sin^2 A + \cos^2 A)+\dfrac34((2 \sin A \cos A)^2) \\ & = 1^3-3 \sin^2 A \cos^2 A(1) + \dfrac{3}{\cancel{4}}(\cancel{4} \sin^2 A \cos^2 A) \\ & = 1-\cancel{3 \sin^2 A \cos^2 A}+\cancel{3 \sin^2 A \cos^2 A} \\ & = 1 \end{aligned}$$Therefore, it is proven that $\sin^6 A + \cos^6 A + \dfrac34 \sin^2 2A = 1.$
Problem Number 37
Prove that $$\sqrt{\sin^4 x + 4 \cos^2 x}-\sqrt{\cos^4 x + 4 \sin^2 x} = \cos 2x.$$
Notice that $-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$, so that $\sin^2 x \le 1$, and similarly $\cos^2 x \le 1$.
The trigonometric identities used are:
$$\begin{aligned} \sin^2 x + \cos^2 x & = 1 && (\text{Pythagorean Identity}) \\ \cos^2 x-\sin^2 x & = \cos 2x && (\text{Double-Angle Identity}) \end{aligned}$$Proof from the left-hand side.
$$\begin{aligned} & \sqrt{\sin^4 x + 4 \cos^2 x}-\sqrt{\cos^4 x + 4 \sin^2 x} \\ & = \sqrt{\sin^4 x + 4 (1-\sin^2 x)}-\sqrt{\cos^4 x + 4 (1-\cos^2 x)} \\ & = \sqrt{\sin^4 x-4 \sin^2 x + 4}-\sqrt{\cos^4 x-4 \cos^2 x + 4} \\ & = \sqrt{(\sin^2 x-2)^2} -\sqrt{(\cos^2 x-2)^2} \\ & = |\sin^2 x-2|-|\cos^2 x-2| \\ & = -(\sin^2 x-2)+(\cos^2 x-2) && (\sin^2 x \le 1; \cos^2 x \le 1) \\ & = \cos^2 x-\sin^2 x = \cos 2x \end{aligned}$$Therefore, it is proven that $$\sqrt{\sin^4 x + 4 \cos^2 x}-\sqrt{\cos^4 x + 4 \sin^2 x} = \cos 2x.$$