Question

Proving trigonometric identities 2(cosx sinx-sinx cos2x)/sin2x =secx

Answer 1

This is not an identity.

$\dfrac{2(\cos(x)\sin(x) - \sin(x)\cos(2x))}{\sin(2x)} \neq \sec(x)$

Check x = π/4, for which we have cos(π/4) = sin(π/4) = 1/√2. Together with sin(2•π/4) = sin(π/2) = 1 and cos(2•π/4) = cos(π/2) = 0, the left side becomes 1, while sec(π/4) = 1/cos(π/4) = √2.

Keeping the left side unchanged, the correct identity would be

$\dfrac{2(\cos(x)\sin(x) - \sin(x)\cos(2x))}{\sin(2x)} = -2\cos(x) + 1 + \sec(x)$

To show this, recall

• sin(2x) = 2 sin(x) cos(x)

• cos(2x) = cos²(x) - sin²(x)

• cos²(x) + sin²(x) = 1

Then we have

$\dfrac{2(\cos(x)\sin(x) - \sin(x)\cos(2x))}{\sin(2x)} = \dfrac{2\cos(x)\sin(x) - 2\sin(x)\cos(2x)}{\sin(2x)} \\\\ = \dfrac{\sin(2x) - 2\sin(x)\cos(2x)}{\sin(2x)} \\\\ = 1 - \dfrac{2\sin(x)\cos(2x)}{\sin(2x)} \\\\ = 1 - \dfrac{2\sin(x)(\cos^2(x) - \sin^2(x))}{2 \sin(x)\cos(x)} \\\\ = 1 - \dfrac{\cos^2(x) - \sin^2(x)}{\cos(x)} \\\\ = 1 - \cos(x) + \dfrac{\sin^2(x)}{\cos(x)} \\\\ = 1 - \cos(x) + \dfrac{1 - \cos^2(x)}{\cos(x)} \\\\ = 1 - \cos(x) + \sec(x) - \cos(x) \\\\ = -2\cos(x) + 1 + \sec(x)$