# Proving Trigonometric Identities

#### Contents

## Proving Trigonometric Identities - Basic

Trigonometric identities are equalities involving trigonometric functions. An example of a trigonometric identity is

\[\sin^2 \theta + \cos^2 \theta = 1.\]

In order to prove trigonometric identities, we generally use other known identities such as Pythagorean Identities.

## Prove that \((1 - \sin x) (1 +\csc x) =\cos x \cot x.\)

We have \[(1 - \sin x) (1 +\csc x)=(1 - \sin x)\left(1 + \frac{1}{\sin x} \right).\]

Expanding the brackets we get

\[\begin{align} 1+ \frac{1}{\sin x} - \sin x -1

&=\frac{1}{\sin x} -\sin x\\ &=\frac{1-\sin^{2} x}{\sin x}\\ &=\frac{\cos^{2} x}{\sin x}\\ &=\cos x \frac{\cos x}{\sin x}\\ &=\cos x \cot x. \ _\square \end{align}\]

## Proving Trigonometric Identities - Intermediate

## Proving Trigonometric Identities - Advanced

## Proving Trigonometric Identities

Proving a trigonometric identity refers to showing that the identity is always true, no matter what value of \( x \) or \( \theta \) is used.

Because it has to hold true for all values of \(x\), we cannot simply substitute in a few values of \(x\) to "show" that they are equal. It is possible that both sides are equal at several values (namely when we solve the equation), and we might falsely think that we have a true identity.

Instead, we have to use logical steps to show that one side of the equation can be transformed to the other side of the equation. Sometimes, we will work separately on each side, till they meet in the middle.

## General Approach

You should be familiar with the various trigonometric identities, like the Reciprocal Trigonometric Functions and the Pythagorean Identities.

There are many different ways to prove an identity. Here are some guidelines in case you get stuck:

1) Work on the side that is more complicated. Try and simplify it.

2) Replace all trigonometric functions with just \( \sin \theta \) and \( \cos \theta \) where possible.

3) Identify algebraic operations like factoring, expanding, distributive property, adding and multiplying fractions. This allows us to simplify the expression further.

4) Use the various Trigonometric Identities. In particular, watch out for the Pythagorean Identity.

5) Work from both sides.

6) Keep an eye on the other side, and work towards it.

7) Consider the "trigonometric conjugate".

## Prove the identity

\[ \frac { \cot \theta } { \csc \theta } = \cos \theta. \]

(Guideline 1) We start from the LHS which is more complicated. Notice that if we were to start from the RHS, it is not clear how we can proceed.

(Guideline 2) We replace \( \cot \theta \) with \( \frac{ \cos \theta } { \sin \theta } \) and \( \csc \theta \) with \( \frac{ 1 } { \sin \theta } \), to obtain:

\[ LHS = \frac{ \cot \theta } { \csc \theta } = \frac{ \frac{ \cos \theta } { \sin \theta } } { \frac{ 1} { \sin \theta }} = \frac{ \cos \theta } { 1} = \cos \theta = RHS.\ _\square \]

The ability to prove trigonometric identities will help with problems like this:

## Prove that

\[ \sin x \cos x \tan x = 1 - \cos^2 x. \]

(Guideline 1) The LHS is more complicated, so let's start with it.

(Guideline 2) Let's replace everything with \( \sin x, \cos x \). We have

\[ LHS = \sin x \cos x \frac{ \sin x } { \cos x } = \sin x \times \sin x = \sin^2 x .\]

(Guideline 4) On the RHS, replace 1 with \( \sin ^2 x + \cos ^2 x \), and we get

\[ RHS = ( \sin^2 x + \cos^2 x ) - \cos^2 x = \sin ^2 x. \]

Thus, \( LHS = \sin^2 x = RHS.\ _\square \).

## Prove the identity

\[ \sec ^4 \theta - \tan^4 \theta = 2 \sec ^2 \theta - 1 .\]

(Guideline 3) We recognize that the LHS can be factorized as

\[ LHS = \sec^4 \theta - \tan ^4 \theta = ( \sec^2 \theta - \tan^2 \theta ) ( \sec^2 \theta + \tan^2 \theta ) = 1 \times ( \sec^2 \theta + \tan ^2 \theta ). \]

(Guideline 2 ... We might want to replace with \( \sin \theta \) and \( \cos \theta \) to proceed further. However, since everything is in just \( \sec \theta \) and \( \tan \theta \), let's keep it that way for now.)

(Guideline 4) We will use the Pythagorean Identity \( \sec^2 \theta = 1 + \tan ^2 \theta \), which gives us

\[ LHS = \sec^2 \theta + \tan ^2 \theta = \sec^2 \theta + ( \sec^2 \theta - 1) = 2 \sec^2 \theta - 1 = RHS.\ _\square \]

## Prove the identity \[ \sin x + \sin x \cot ^2 x = \csc x. \]

(Guideline 3) The LHS can be factorized as

\[ LHS = \sin x ( 1 + \cot^2 x ). \]

(Guideline 4) We recognize that \( 1 + \cot^2 x = \csc ^2 x \), so we get

\[ LHS = \sin x ( \csc ^2 x ). \]

(Guideline 2) Replace with \( \sin x \), we get

\[ LHS = \sin x \times \frac { 1} { \sin ^2 x } = \frac{1} { \sin x } = \csc x = RHS.\ _\square \]

## Prove the identity

\[ \frac{ \sin \theta } { 1 - \cos \theta } + \frac{ 1 - \cos \theta } { \sin \theta } = 2 \csc \theta .\]

(Guideline 2, 5) Since the LHS is already in terms of \( \sin \theta \) and \( \cos \theta \), we simplify the RHS to

\[ RHS = \frac{ 2 } { \sin \theta } . \]

(Guideline 1) The LHS is certainly more complicated. Let's try and see what we could do with it. If we blindly combined fractions (Guideline 3), we will get

\[ LHS = \frac{ \sin^2 \theta + ( 1 - \cos \theta )^2 } { \sin \theta ( 1 - \cos \theta) }. \]

This is beginning to look good, because we have \( \sin \theta \) in the denominator. We need to figure out how to proceed further. Let's investigate the numerator:

\[ \sin ^2\theta + ( 1 - \cos\theta)^2 = \sin ^2 \theta + 1 -- 2 \cos \theta + \cos ^2 \theta = 1 - 2 \cos \theta + 1 = 2 - 2 \cos \theta . \]

Ah-ha! There is a much simplier way to write the numerator, and the LHS becomes

\[ LHS = \frac{ 2 - 2 \cos \theta } { \sin \theta ( 1 - \cos \theta )} = \frac{ 2 } { \sin \theta } = RHS.\ _\square \]

## Using the sum and product formulas

## Prove the identity

\[ \tan \theta + \cot \theta = \frac{ 2} { \sin 2 \theta }. \]

(Guideline 2) Replacing \( \tan \theta \) with \( \frac{ \sin \theta } { \cos \theta } \) and \( \cot \theta \) with \( \frac { \cos \theta } { \sin \theta } \),

\[LHS = \frac{ \sin \theta } { \cos \theta } + \frac { \cos \theta } + \frac { \cos \theta } { \sin \theta } = \frac{ \sin ^2 \theta + \cos ^2 \theta } { \sin \theta \cos \theta } = \frac{ 1} { \sin \theta \cos \theta }. \]

(Guideline 5) We are now stuck as this is very simple. Let's look at the RHS. Replacing \( \sin 2 \theta\) with \( 2 \sin \theta \cos \theta \), we get

\[ RHS =\frac {2} { \sin 2 \theta } = \frac{ 2 } { 2 \sin \theta \cos \theta } = \frac{ 1 } { \sin \theta \cos \theta } . \]

Thus, we have

\[ LHS = \frac{ 1}{ \sin \theta \cos \theta } =RHS.\ _\square \]

## Prove the identity

\[ \frac{ \sin x - \cos x } { \sin x + \cos x } = - \frac{ \cos 2x } { 1 + \sin 2 x }. \]

(Guideline 6) This identity is hard to attack directly. Let's think about what the target is, and compare the sides to each other.

(Guideline 4) Let's replace \( \cos 2x \). We know that there are 3 choices, since

\[ \cos 2 x = 2 \cos^2 x - 1 = \cos^2 x - \sin ^2 x = 1 - 2 \sin^2x . \]

Which one should we choose? Notice that if we took \( \cos^2 x - \sin^2 x \), that allows us to factorize it as \( ( \cos x - \sin x ) ( \cos x + \sin x ) \), which is close to the LHS. Thus, we have

\[ RHS = - \frac{ \cos^2 x - \sin^2 x } { 1 + \sin 2x } = \frac{ -(\cos x - \sin x ) ( \cos x + \sin x ) } { 1 + \sin 2x }. \]

We now have the \( \sin x - \cos x \) which is in the numerator of the LHS, so we know that the denominator of the RHS must be something similar. In fact, if this is an identity, it must be equal to \( (\sin x + \cos x)^ 2 \). We can verify that

\[ 1 + \sin 2 x = 1 + 2 \sin x \cos x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = ( \sin x + \cos x ) ^2. \]

Thus, this give us

\[ RHS = \frac{ - ( \cos x - \sin x ) ( \cos x + \sin x ) } { ( \sin x + \cos x ) ( \sin x + \cos x ) } = \frac{ \sin x - \cos x } { \sin x + \cos x } = LHS.\ _\square \]

## Prove the identity

\[ \frac{ \sin ( A + B) + \sin (A - B) } { \sin (A + B) - \sin (A - B ) } = \frac{ \tan A } { \tan B }. \]

Using the sum and difference formulas, we get

\[\begin{align} LHS &= \frac{ (\sin A \cos B + \cos A \sin B) + (\sin A \cos B - \cos A \sin B)} { (\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B) } \\ &= \frac { 2 \sin A \cos B } { 2 \cos A \sin B } \\ &= \frac{ \sin A \cos B } { \cos A \sin B } . \end{align}\]

Simplifying the RHS in terms of sin and cos, we get

\[ RHS = \frac{ \frac{ \sin A } { \cos A } } { \frac { \sin B } { \cos B } } = \frac { \sin A \cos B } { \cos A \sin B }. \]

Hence, \( LHS = RHS.\ _\square\)

## Prove the identity

\[\sin3\theta=-4\sin^3\theta+3\sin\theta.\]

Consider \(3\theta\) as \(2\theta+\theta\). Then applying the angle sum formula for sine gives

\[LHS=\sin3\theta=\sin(2\theta+\theta)=\sin2\theta\cos\theta+\cos2\theta\sin\theta.\]

Now substituting \(\sin2\theta=2\sin\theta\cos\theta\) and \(\cos2\theta=2\cos^2\theta-1\) into the equation gives

\[\begin{align} \sin2\theta\cos\theta+\cos2\theta\sin\theta&=2\sin\theta\cos^2\theta+(2\cos^2\theta-1)\sin\theta\\ &=4\sin\theta\cos^2\theta-\sin\theta. \end{align}\]

Replace \(\cos^2\theta\) with \(1-\sin^2\theta\). Then we have

\[\begin{align} 4\sin\theta\cos^2\theta-\sin\theta&=4\sin\theta(1-\sin^2\theta)-\sin\theta\\ &=-4\sin^3\theta+3\sin\theta=RHS.\ _\square \end{align}\]

## Prove the identity

\[\cot\theta-\cot2\theta=\frac{1}{\sin2\theta}.\]

Since \(\cot\theta=\frac{\cos\theta}{\sin\theta}\), we have

\[\cot\theta-\cot2\theta=\frac{\cos\theta}{\sin\theta}-\frac{\cos2\theta}{\sin2\theta}.\]

Since \(\sin2\theta=2\sin\theta\cos\theta\), the fractions above can be reduced to a common denominator, which is \(2\sin\theta\cos\theta\), as shown below:

\[\frac{\cos\theta}{\sin\theta}-\frac{\cos2\theta}{\sin2\theta}=\frac{\cos\theta}{\sin\theta}-\frac{\cos2\theta}{2\sin\theta\cos\theta}=\frac{2\cos^2\theta-\cos2\theta}{2\sin\theta\cos\theta}.\]

Applying the formula \(\cos2\theta=2\cos^2\theta-1\) to this leads to the RHS:

\[\frac{2\cos^2\theta-\cos2\theta}{2\sin\theta\cos\theta}=\frac{2\cos^2\theta-(2\cos^2\theta-1)}{2\sin\theta\cos\theta}=\frac{1}{2\sin\theta\cos\theta}=\frac{1}{\sin2\theta}.\ _\square\]

## Prove the identity

\[\frac{\sin^2A-\sin^2B}{\sin^2(A+B)}=\frac{\tan A-\tan B}{\tan A + \tan B}.\]

At first it might be tempting to factorize \(\sin^2 A-\sin^2 B\) into \((\sin A+\sin B)(\sin A-\sin B)\), but you will soon realize that this leads to nothing. In order to move on, we should apply the power reduction formula, which gives

\[\frac{\sin^2A-\sin^2B}{\sin^2(A+B)}=\frac{\frac{1-\cos2A}{2}-\frac{1-\cos2B}{2}}{\sin^2(A+B)}=\frac{\cos2B-\cos2A}{2\sin^2(A+B)}.\]

Now we can use the sum and product formula for the numerator, as shown below:

\[\begin{align} \frac{\cos2B-\cos2A}{2\sin^2(A+B)}&=\frac{-2\sin(B+A)\sin(B-A)}{2\sin^2(A+B)}\\ &=\frac{\sin(A+B)\sin(A-B)}{\sin^2(A+B)}\\ &=\frac{\sin(A-B)}{\sin(A+B)}. \end{align}\]

The next step is simply applying the angle sum formula to the numerator and denominator. Then we have

\[\frac{\sin(A-B)}{\sin(A+B)}=\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\cos A\sin B}.\]

Now, we divide both the numerator and denominator by \(\cos A\cos B\) in order to obtain tangents as follows:

\[\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\cos A\sin B}=\frac{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}=\frac{\tan A-\tan B}{\tan A+\tan B}.\ _\square\]

[Another Solution]

The solution above might be quite difficult to think of. If you find yourself stuck while struggling to organize the LHS, switching over and tackling the RHS might work. Just try converting the tangents into sines and cosines by using the formula \(\tan=\frac{\sin\theta}{\cos\theta}\), and solve the problem in the opposite direction of the first solution, as shown below:

\[\begin{align} \frac{\tan A-\tan B}{\tan A + \tan B}&=\frac{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}\\ &=\frac{\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}}{\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}}\\ &=\frac{\sin(A-B)}{\sin(A+B)}\\ &=\frac{\sin(A-B)\sin(A+B)}{\sin^2(A+B)}\\ &=-\frac{1}{2}\cdot\frac{\cos2A-\cos2B}{\sin^2(A+B)}\\ &=-\frac{1}{2}\cdot\frac{1-2\sin^2A-(1-2\sin^2B)}{\sin^2(A+B)}\\ &=\frac{\sin^2A-\sin^2B}{\sin^2(A+B)}\\ &=LHS.\ _\square \end{align}\]

## Conditional Identities

Prove that if \( \alpha + \beta + \gamma = \pi \), then\[ \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \times \tan \beta \times \tan \gamma. \]

Since \( \alpha = \pi - \beta - \gamma\), we have

\[ \begin{align}

\tan \alpha & = \tan ( \pi - \beta - \gamma) \\ &= - \tan ( \beta + \gamma) \\ & = - \frac{ \tan \beta + \tan \gamma } { 1 - \tan \beta \tan \gamma} \\ &= \frac{ \tan \beta + \tan \gamma } { \tan \beta \tan \gamma - 1 }. \end{align} \]Cross multiplying, we get that

\[ \tan \alpha \tan \beta \tan \gamma - \tan \alpha = \tan \beta + \tan \gamma. \]

If we add \( \tan \alpha \) to both sides, we get the identity that we wanted to show.\(\ _\square \)

Prove the identity\[\large \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}. \]

If \(n\) is even, then \[\cos\left(\frac{\pi}{2}\right)=0\] appears in the product when \(k=\frac{n}{2}\) and \[\sin\left(\frac{n\pi}{2}\right)=0.\] If \(n\) is odd, then combining \[\begin{align} \lim_{z=1}\frac{z^n+1}{z+1}&=1 &\qquad (1) \\ \frac{z^n+1}{z+1}&=\prod_{k=1}^{n-1}(z+e^{2\pi ik/n}) &\qquad (2) \\ 1+e^{i2k\pi/n}&=2\cos(k\pi/n)e^{ik\pi/n} &\qquad (3) \end{align} \] and noting that \[\displaystyle\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}\] gives \[\displaystyle\prod_{k=1}^{n-1}e^{ik\pi/n}=(-1)^{(n-1)/2}.\] This matches the sign of \[\sin(\pi n/2)\] and yields \[\begin{align} 2^{n-1}\prod_{k=1}^{n-1}\cos(k\pi/n) &=(-1)^{(n-1)/2}\\ &=\sin(\pi n/2).\ _\square \end{align} \]

**Cite as:**Proving Trigonometric Identities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/proving-trigonometric-identities/