Question

Proving trigonometric identities

2(cosx sinx-sinx cos2x)/sin2x =secx

Answer 1

This is not an identity.

`(2(\cos(x)\sin(x) - \sin(x)\cos(2x)))/(\sin(2x)) \\eq \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

`(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

`(2(\cos(x)\sin(x) - \sin(x)\cos(2x)))/(\sin(2x)) = (2\cos(x)\sin(x) - 2\sin(x)\cos(2x))/(\sin(2x)) \\\\ = (\sin(2x) - 2\sin(x)\cos(2x))/(\sin(2x)) \\\\ = 1 - (2\sin(x)\cos(2x))/(\sin(2x)) \\\\ = 1 - (2\sin(x)(\cos^2(x) - \sin^2(x)))/(2 \sin(x)\cos(x)) \\\\ = 1 - (\cos^2(x) - \sin^2(x))/(\cos(x)) \\\\ = 1 - \cos(x) + (\sin^2(x))/(\cos(x)) \\\\ = 1 - \cos(x) + (1 - \cos^2(x))/(\cos(x)) \\\\ = 1 - \cos(x) + \sec(x) - \cos(x) \\\\ = -2\cos(x) + 1 + \sec(x)`