+0

# Simplify trig sum

+1
295
2

I am working on a sum, but get stuck...

$$Cos(90+x) \over{Cos(360-x)Tan(180-x)Cos480}$$

$$-Sinx \over{Cosx.Tanx.Cos120}$$

Is this right so far?...not sure how to proceed with this...please help

May 31, 2019

### 2+0 Answers

#1
+12212
+1

I am working on a sum, but get stuck...

May 31, 2019
edited by Omi67  May 31, 2019
#2
+8962
+3

$$\begin{array}{} \phantom{=\qquad}\frac{\cos(90°+x)}{\cos(360°-x)\tan(180°-x)\cos(480°)}&\\~\\ =\qquad\frac{-\sin(x)}{\cos(360°-x)\tan(180°-x)\cos(480°)}&\qquad\text{because}\qquad \cos(90°+x)=-\sin (x) \\~\\ =\qquad\frac{-\sin(x)}{\cos(-x)\tan(180°-x)\cos(480°)}&\qquad\text{because}\qquad \cos(360°-x)=\cos(-x+360°)=\cos(-x) \\~\\ =\qquad\frac{-\sin(x)}{\cos(x)\tan(180°-x)\cos(480°)}&\qquad\text{because cos(x) is an even function,}\quad \cos(-x)=\cos(x) \\~\\ =\qquad\frac{-\sin(x)}{\cos(x)(-\tan(x))\cos(480°)}&\qquad\text{because}\qquad \tan(180°-x)=-\tan(x)\\~\\ =\qquad\frac{\sin(x)}{\cos(x)\tan(x)\cos(480°)}&\\~\\ =\qquad\frac{\sin(x)}{\cos(x)\tan(x)\cos(120°)}&\qquad\text{because}\qquad \cos(480°)=\cos(480°-360°)=\cos(120°) \end{array}$$

So far the only difference is because  tan(180° - x)  =  -tan(x)

$$\begin{array}{} \phantom{=\qquad}\frac{\sin(x)}{\cos(x)\tan(x)\cos(120°)}&\\~\\ =\qquad\frac{\sin(x)}{\cos(x)}\cdot\frac{1}{\tan(x)\cos(120°)}&\\~\\ =\qquad\tan(x)\cdot\frac{1}{\tan(x)\cos(120°)}&\qquad\text{because}\qquad \frac{\sin(x)}{\cos(x)}=\tan(x)\\~\\ =\qquad\frac{1}{\cos(120°)}&\\~\\ =\qquad\frac{1}{(-\frac12)}&\qquad\text{because}\qquad \cos(120°)=-\frac12\\~\\ =\qquad-2& \end{array}$$

May 31, 2019