\(\sin(A+B)\)=\(\sin A\cos B+\cos A\sin B\)
Put A=B, then
\(\sin(A+A)\)=\(\sin A\cos A+\cos A\sin A\)
i.e, \(\sin2A=2\sin A\cos A\)
\(\cos2A\)=\(\cos(A+A)\)
=\(\cos A\cos A-\sin A\sin A\)
=\(\cos^{2}A-\sin^{2}A\)
=1-\(\sin^{2}A-\sin^{2}A\)
=1-\(2\sin^{2}A\)
\(\cos2A\)=\(\cos(A+A)\)
=\(\cos A\cos A-\sin A\sin A\)
=\(\cos^{2}A-1+\cos^{2}A\)
=2\(\cos^{2}A-1\)
\(2\cos^{2}A=1+\cos2A\)...(1)
\(2\sin^{2}A=1-\cos2A\)...(2)
Equation (1) divide by equation (2), we get
\(\frac{2\sin^{2}A}{2\cos^{2}A}=\frac{1-\cos2A}{1+\cos2A}\)
\(\Rightarrow\) \(\tan^{2}A=\frac{1-\cos2A}{1+\cos2A}\)
\(\Rightarrow\) \(\frac{2\sin^{2}A}{2\cos^{2}A}=\frac{1-\cos2A}{1+\cos2A}\)
\(\Rightarrow\) \(\tan^{2}A=\frac{1-\cos2A}{1+\cos2A}\)
\(\Rightarrow\) \(1-\cos2A=\tan^{2}A(1+\cos2A)\)
\(\Rightarrow\) \(1-\cos2A=\tan^{2}A+\tan^{2}A\cos2A\)
\(\Rightarrow\) \(1-\tan^{2}A=(1+\tan^{2}A)\cos2A\)
\(\Rightarrow\) \(\cos2A=\frac{1-\tan^{2}A}{1+\tan^{2}A}\)
\(\sin2A=2\sin A\cos A\)
=\(2\frac{\sin A}{\cos A}\times\frac{1}{\sec^{2}A}\)
=\(\frac{2\tan A}{1+\tan^{2}A}\)
\(\tan2A=\tan(A+A)\)
=\(\frac{\tan A+\tan A}{1-\tan A\tan A}\)
=\(\frac{2\tan A}{1-\tan^{2}A}\)
\(\sin3A\)=\(\sin(2A+A)\)
=\(\sin2A\cos A+ \cos2A\sin A\)
=2\(\sin A\cos A\cos A+ (1-2\sin^{2}A)\sin A\)
=2\(\sin A(1-\sin^{2}A)+\sin A-2\sin^{3}A\)
=2\(\sin A-2\sin^{3}A+\sin A-2\sin^{3}A\)
=3\(\sin A-4\sin^{3}A\)
\(\cos3A\)=\(\cos(2A+A)\)
=\(\cos2A\cos A- \sin2A\sin A\)
=\((2\cos^{2}A-1)\cos A-2\sin A\cos A\sin A\)
=\(2\cos^{3}A-\cos A-2\cos A(1-\cos^{2}A)\)
=\(2\cos^{3}A-\cos A-2\cos A+2\cos^{3}A\)
=\(4\cos^{3}A-3\cos A\)