Question

Cos(x+y)cos(x-y)=cos^2(x)-sin^2(y) Verify the identity

Answer 1

Answer: Verified.

Step-by-step explanation:  The given identity involving trigonometric ratios is

$\cos(x+y)\cos(x-y)=\cos^2x-\sin^2y.$

We have

$L.H.S.\\\\=\cos(x+y)\cos(x-y)\\\\=\left(\cos x\cos y-\sin x\sin y\right)\left(\cos x\cos y+\sin x\sin y\right)\\\\=\left(\cos x\cos y\right)^2-\left(\sin x\sin y\right)^2\\\\=\cos^2x\cos^2y-\sin^2x\sin^2y\\\\=\cos^2x(1-\sin^2y)-(1-\cos^2x)\sin^2y\\\\=\cos^2x-\cos^2x\sin^2y-\sin^2y+\cos^2x\sin^2y\\\\=\cos^2x-\sin^2y\\\\=R.H.S.$

Therefore, L.H.S. = R.H.S.

Thus, the given identity is verified.

Answer 2

Answer:

cos(x+y)sin(x-y) = cos²x-sin²y ⇒ L.H.S = R.H.S

Step-by-step explanation:

L.H.S:

∵ cos(x+y)cos(x-y) = (cosxcosy - sinxsiny)(cosxcosy + sinxsiny)

Multiply the brackets ⇒ the answer is difference of two squares

∴ = cos²xcos²y - sin²xsin²y

∵ cos²y = 1 - sin²y and sin²x = 1 - cos²x

∴ cos²xcos²y - sin²xsin²y = cos²x (1 - sin²y) - (1  -cos²x) sin²y

∴ = cos²x - cos²xsin²y - sin²y + cos²xsin²y

∵ -cos²x sin²y will cancel cos²xsin²y ⇒ same terms with different sign

∴ = cos²x - sin²y = R.H.S