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Eleven Standard >> Compound angles | Introduction

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Introduction of Compound angles

 

Compound angles

In the above picture, the circle is a unit circle whose radius OP is unit and the center that is origin is O.

Let \(\angle POQ=A\)
      \(\angle POR=B\)                                                        
So \(\angle POS=A-B\)                                                         
   \(\angle QOR=\angle POQ-\angle POR=A-B\) 
Length of \(\overline{QR}\)=Length of \(\overline{PS}\) 
 \(\triangle ORT\ RT\perp OP\)  
 \(\frac{OT}{OR}=\cos B,\ \frac{RT}{OR}=\sin B\) 
 \(\therefore\ OT=\cos B,\ RT=\sin B\) 
     [\(\because\ OR=unit\)]
 \( \overline{QR}=\overline{PS} \)                                     
\(\Rightarrow \sqrt{(\cos A-\cos B)^{2}+(\sin A-\sin B)^{2}}\)
\(\Rightarrow \sqrt{((\cos (A-B)-0)^{2}+(\sin (A-B)-0)^{2}}\)                                                                                      

By squaring both sides we get     
\(\Rightarrow\ \cos^{2}A+\cos^{2}B-2\cos A\cos B+\sin^{2}A+\sin^{2}B-2\sin A\sin B\) =\(\cos^{2}(A-B)+1-2\cos (A-B)+\sin^{2}(A-B)\)
\(\Rightarrow\ \sin^{2}A+\cos^{2}A+\sin^{2}B+\cos^{2}B-2(\cos A\cos B+2\sin A\sin B)\)=\(\sin^{2}(A-B)+\cos^{2}(A-B)+1-2\cos (A-B)\)
\(\Rightarrow 1+1-2(\cos A\cos B+2\sin A\sin B)\)=\(1+1-2\cos (A-B)\)
\(\Rightarrow\ 2(\cos A\cos B+2\sin A\sin B)\)=\(2\cos (A-B)\) 
\(\Rightarrow\ \cos (A-B)\)=\(2(\cos A\cos B+2\sin A\sin B)\)  

Find the value of \(\cos 15^{0}\)
\(\cos 15^{0}=\cos (45^{0}-30^{0})\)
          =\(\cos 45^{0}\cos 30^{0}+\cos 45^{0}\cos 30^{0}\)
          =\(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}\)
          =\(\frac{\sqrt{3}+1}{2\sqrt{2}}\)

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