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