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10.04 Verifying trig identities


We have discussed several different trigonometric identities. But there are actually much more to be discovered. In fact, we can use the basic trigonometric identities below to verify many more trig identities.

Basic trigonometric identities
$\tan\theta\equiv\frac{\sin\theta}{\cos\theta}$tanθsinθcosθ $\csc\theta\equiv\frac{1}{\sin\theta}$cscθ1sinθ $\sin^2\theta+\cos^2\theta\equiv1$sin2θ+cos2θ1
$\cot\theta\equiv\frac{\cos\theta}{\sin\theta}$cotθcosθsinθ $\sec\theta\equiv\frac{1}{\cos\theta}$secθ1cosθ $\tan^2\theta+1\equiv\sec^2\theta$tan2θ+1sec2θ
  $\cot\theta\equiv\frac{1}{\tan\theta}$cotθ1tanθ $1+\cot^2\theta\equiv\csc^2\theta$1+cot2θcsc2θ

Let's walk through some examples.

Worked examples

Question 1

Verify the identity $\frac{\sin x}{\cos x\tan x}=1$sinxcosxtanx=1.

Think: Since verifying an identity is a proof, we cannot manipulate both sides of the equation at once. Instead, we must work with one side of the equation and show that algebraic manipulation leads to the other side.

Do: Manipulate the left-hand side of the identity:

$\frac{\sin x}{\cos x\tan x}$sinxcosxtanx $=$= $\frac{\sin x}{\cos x\tan x}$sinxcosxtanx

Start with the left-hand side

  $=$= $\frac{\sin x}{\cos x\times\frac{\sin x}{\cos x}}$sinxcosx×sinxcosx

Substitute $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx

  $=$= $\frac{\sin x}{\frac{\cos x}{1}\times\frac{\sin x}{\cos x}}$sinxcosx1×sinxcosx

Multiply the fractions

  $=$= $\frac{\sin x}{\sin x}$sinxsinx

Cancel the common factor of $\cos x$cosx

  $=$= $1$1

Simplify the expression


Question 2

Verify that $1+\tan^2\theta=\sec^2\theta$1+tan2θ=sec2θ using other fundamental identities.

Think: It's often useful to rewrite the tangent and reciprocal functions in terms of the sine and cosine functions. Doing so may require adding fractions, so it's good to keep the rules of fractions in mind.

Do: Perform the following manipulations, starting with the left-hand side.

$1+\tan^2\theta$1+tan2θ $=$= $1+\tan^2\theta$1+tan2θ

Start with the left-hand side

  $=$= $1+\frac{\sin^2\theta}{\cos^2\theta}$1+sin2θcos2θ

Use the fundamental identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ

  $=$= $\frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}$cos2θcos2θ+sin2θcos2θ

Create a common denominator by multiplying $1$1 by $\frac{\cos^2\theta}{\cos^2\theta}$cos2θcos2θ

  $=$= $\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}$cos2θ+sin2θcos2θ

Add the numerators

  $=$= $\frac{1}{\cos^2\theta}$1cos2θ

Use the fundamental identity $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1

  $=$= $\sec^2\theta$sec2θ

Use the fundamental identity $\sec^2\theta=\frac{1}{\cos^2\theta}$sec2θ=1cos2θ


Reflect: Notice how we did not use the identity $1+\tan^2\theta=\sec^2\theta$1+tan2θ=sec2θ itself in the proof. Instead, we must verify an identity by using other identities that have already been verified.


Question 3

Verify the identity $\frac{\sin\theta}{1-\cos\theta}=\frac{1+\cos\theta}{\sin\theta}$sinθ1cosθ=1+cosθsinθ.

Think: If a fractional expression contains $1+\cos\theta$1+cosθ, multiplying both the numerator and denominator by $1-\cos\theta$1cosθ would give $1-\cos^2\left(\theta\right)$1cos2(θ), which we could simplify using one of the trigonometric identities.

Do: Starting with the left-hand side, perform the following algebraic maniupulations:

$\frac{\sin\theta}{1-\cos\theta}$sinθ1cosθ $=$= $\frac{\sin\theta}{1-\cos\theta}$sinθ1cosθ

Begin with the left-hand side.

  $=$= $\frac{\sin\theta}{1-\cos\theta}\times\frac{1+\cos\theta}{1+\cos\theta}$sinθ1cosθ×1+cosθ1+cosθ

Multiply by $1=\frac{1+\cos\theta}{1+\cos\theta}$1=1+cosθ1+cosθ

  $=$= $\frac{\sin\theta\left(1+\cos\theta\right)}{1-\cos^2\left(\theta\right)}$sinθ(1+cosθ)1cos2(θ)

Use the distributive property in the denominator $\left(A-B\right)\left(A+B\right)=A^2-B^2$(AB)(A+B)=A2B2

  $=$= $\frac{\sin\theta\left(1+\cos\theta\right)}{\sin^2\left(\theta\right)}$sinθ(1+cosθ)sin2(θ)

Rewrite the denominator using the identity $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1

  $=$= $\frac{1+\cos\theta}{\sin\theta}$1+cosθsinθ

Cancel out the common factor of $\sin\theta$sinθ



Reflect: We could have also verified the identity by transforming the right-hand side into the left-hand side.



If an assumed identity contains fractions, it is not sufficient to multiply both sides of the equation by the least common denominator. Instead, we should focus on each side independently.


Practice questions

Question 4

Prove that $\frac{\tan x\cos x}{\sin x}=1$tanxcosxsinx=1.

Question 5

Prove the identity $2\cos^2\left(\theta\right)-3=-1-2\sin^2\left(\theta\right)$2cos2(θ)3=12sin2(θ).

Question 6

Prove the identity $\sin\theta\left(1+\tan\theta\right)+\cos\theta\left(1+\cot\theta\right)=\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}$sinθ(1+tanθ)+cosθ(1+cotθ)=sinθ+cosθsinθcosθ.

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