Question

Please prove the following trigonometric identity.


`\cos^2(5x)-\sin^2(5x)=\cos(10x)`

Answer 1

cos²A - sin²A = cos2A, this is a general indentity where A can have any random value x/2, y/3, 1/8, 8, 9 etc.

Replacing A with 5x:

=> cos²(5x) - sin²(5x) = cos²(10x)

Moreover,

cos(A + B) = cosAcosB - sinAsinB.

When A = B,

cos(A + A) = cosAcosA - sinAsinA

cos2A = cos²A - sin²A

You can do the same with 5x.

cos(10x) = cos(5x + 5x)

cos(10x) = cos(5x)cos(5x) - sin(5x)sin(5x)

cos(10x) = cos²(5x) - sin²(5x)

Answer 2

See Below.

Explanation:

We want to verify the trigonometric identity:

`\cos^2(5x)-\sin^2(5x)=\cos(10x)`

For simplicity, we can let u = 5x. Therefore, by substitution, we acquire:

`\cos^2(u)-\sin^2(u)`

Recall the double-angle identity formula(s) for cosine:

`\displaystyle \begin{aligned} \cos(2x)&=\cos^2(x)-\sin^2(x)\\&=2\cos^2(x)-1\\&=1-2\sin^2(x)\end{aligned}`

Therefore, our identity becomes:

`=\cos(2u)`

Back-substitute:

`=\cos(2(5x))=\cos(10x)\stackrel{\checkmark}{=}\cos(10x)`

Hence proven.