Question #53373

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

Expert's answer

Answer on Question #53373 – Math – Trigonometry

Prove that


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


where xx is a constant.

Solution

We'll use next trigonometric identities


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


Thus we have


3sin⁡x+sin⁡(2x)1+3cos⁡x+cos⁡(2x)=3sin⁡x+2sin⁡(x)cos⁡(x)1+3cos⁡x+cos⁡2x−sin⁡2x==sin⁡(x)(3+2cos⁡(x))1+3cos⁡x+cos⁡2x−(1−cos⁡2x)=sin⁡(x)(3+2cos⁡(x))1+3cos⁡x+cos⁡2x−1+cos⁡2x==sin⁡(x)(3+2cos⁡(x))3cos⁡x+2cos⁡2x=sin⁡(x)(3+2cos⁡(x))cos⁡(x)(3+2cos⁡(x))=sin⁡(x)cos⁡(x)=tan⁡x.\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


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


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