Question

Please help prove that (3sinx+2sin2x)/(1+3cosx+cos2x)=tanx where x is a constant......

Answer 1

Answer on Question #53373 – Math – Trigonometry

Prove that

$$\frac {3 \sin x + \sin (2 x)}{1 + 3 \cos x + \cos (2 x)} = \tan x$$

where $x$ is a constant.

Solution

We'll use next trigonometric identities

$$\sin (2 x) = 2 \sin (x) \cos (x);$$ $$\cos (2 x) = \cos^ {2} x - \sin^ {2} x;$$ $$\sin^ {2} x = 1 - \cos^ {2} x.$$

Thus we have

$$\begin{array}{l} \frac {3 \sin x + \sin (2 x)}{1 + 3 \cos x + \cos (2 x)} = \frac {3 \sin x + 2 \sin (x) \cos (x)}{1 + 3 \cos x + \cos^ {2} x - \sin^ {2} x} = \\ = \frac {\sin (x) (3 + 2 \cos (x))}{1 + 3 \cos x + \cos^ {2} x - (1 - \cos^ {2} x)} = \frac {\sin (x) (3 + 2 \cos (x))}{1 + 3 \cos x + \cos^ {2} x - 1 + \cos^ {2} x} = \\ = \frac {\sin (x) (3 + 2 \cos (x))}{3 \cos x + 2 \cos^ {2} x} = \frac {\sin (x) (3 + 2 \cos (x))}{\cos (x) (3 + 2 \cos (x))} = \frac {\sin (x)}{\cos (x)} = \tan x. \end{array}$$

So we proved

$$\frac {3 \sin x + \sin (2 x)}{1 + 3 \cos x + \cos (2 x)} = \tan x.$$

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