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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
 #1
avatar+10536 
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I am working on a sum, but get stuck...

laugh

 May 31, 2019
edited by Omi67  May 31, 2019
 #2
avatar+8756 
+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)

 

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

 

Check: https://www.wolframalpha.com/input/?i=cos(pi%2F2%2Bx)%2F(cos(2pi-x)tan(pi-x)cos(8pi%2F3))

 May 31, 2019

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