Question

cosx-cos3x/sin3x-sinx = tan2x

Show all your working please

Answer 1

\[\frac{ cosx-\cos3x }{ \sin3x-sinx3 } = \tan2x\]

Answer 2

Use two trigonometric identities for sin 3x and cos 3x.
\[\sin(3x) = 3 \sin x - 4 \sin^3 x\]
\[\cos(3x) = 4 \cos^3x - 3 \cos x\]
Substitute in your formula
\[\frac{ \sin x-\cos 3x }{ \sin 3x-\sin x } \]
\[ = \frac{ \sin x - 4 \cos^3x + 3 \cos x } { 3 \sin x - 4 \sin^3 x - \sin x } \]
\[ = \frac{ 4 \cos x (1- \cos^2x) } {2 \sin x (1-2 \sin^2x)} \]
From trigonometric identities \[ 1= \sin^2 x + \cos^2x \]
and \[ \cos 2x = 1-2 \sin^2x \]
we get
\[ = \frac{4 \cos x \sin^2x} {2 \sin x \cos 2x} \]
From trigonometric identity
\[ \sin 2x = 2 \sin x \cos x\]
we get
\[ = \frac {2 \sin x \sin 2x} {2 \sin x \cos 2x} = \tan 2x \]

Answer 3

Thank you very much for the thorough explanation! :)

Answer 4

@Steve.Irwin You are welcome:)