6.05 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.
| $\tan\theta\equiv\frac{\sin\theta}{\cos\theta}$tanθ≡sinθcosθ | $\csc\theta\equiv\frac{1}{\sin\theta}$cscθ≡1sinθ | $\sin^2\theta+\cos^2\theta\equiv1$sin2θ+cos2θ≡1 |
| $\cot\theta\equiv\frac{\cos\theta}{\sin\theta}$cotθ≡cosθsinθ | $\sec\theta\equiv\frac{1}{\cos\theta}$secθ≡1cosθ | $\tan^2\theta+1\equiv\sec^2\theta$tan2θ+1≡sec2θ |
| $\cot\theta\equiv\frac{1}{\tan\theta}$cotθ≡1tanθ | $1+\cot^2\theta\equiv\csc^2\theta$1+cot2θ≡csc2θ |
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 |
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| $=$= | $\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 $\frac{\sin\theta}{1-\cos\theta}=\frac{1+\cos\theta}{\sin\theta}$sinθ1−cosθ=1+cosθsinθ.
Think: If a fractional expression contains $1+\cos\theta$1+cosθ, multiplying both the numerator and denominator by $1-\cos\theta$1−cosθ would give $1-\cos^2\left(\theta\right)$1−cos2(θ), which we could simplify using one of the trigonometric identities.
Do: Starting with the left-hand side, perform the following algebraic maniupulations:
| $\frac{\sin\theta}{1-\cos\theta}$sinθ1−cosθ | $=$= | $\frac{\sin\theta}{1-\cos\theta}$sinθ1−cosθ |
Begin with the left-hand side. |
| $=$= | $\frac{\sin\theta}{1-\cos\theta}\times\frac{1+\cos\theta}{1+\cos\theta}$sinθ1−cosθ×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)1−cos2(θ) |
Use the distributive property in the denominator $\left(A-B\right)\left(A+B\right)=A^2-B^2$(A−B)(A+B)=A2−B2 |
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| $=$= | $\frac{\sin\theta\left(1+\cos\theta\right)}{\sin^2\left(\theta\right)}$sinθ(1+cosθ)sin2(θ) |
Rewrite the denominator using the identity $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1 |
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| $=$= | $\frac{1+\cos\theta}{\sin\theta}$1+cosθsinθ |
Cancel out the common factor of $\sin\theta$sinθ
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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=−1−2sin2(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.