Question

write the formula for $\cos 3A$.

Answer 1

Formula for $\cos 3A$.

It is known that,

$\cos \left(a+b\right)=\cos \left(a\right)\cos \left(b\right)-\sin \left(a\right)\sin \left(b\right)$

Write the formula for $\cos 3A$.

$\begin{array}{rcl}\cos \left(3A\right)& =& \cos \left(2A+A\right)\\ & \Rightarrow & \cos \left(3A\right)=\cos 2A\cos A-\sin 2A\sin A\left[\because \cos \left(a+b\right)=\cos \left(a\right)\cos \left(b\right)-\sin \left(a\right)\sin \left(b\right)\right]\\ & \Rightarrow & \cos \left(3A\right)=\left(2{\cos }^{2}A-1\right)\cos A-2\cos A\sin A\sin A\left[\because \cos 2A=2{\cos }^{2}A-1\right]\\ & \Rightarrow & \cos \left(3A\right)=2{\cos }^{3}A-\cos A-2{\sin }^{2}A\cos A\\ & \Rightarrow & \cos \left(3A\right)=2{\cos }^{3}A-\cos A-2\left(1-{\cos }^{2}A\right)\cos A\left[\because {\cos }^{2}A=1-{\sin }^{2}A\right]\\ & \Rightarrow & \cos \left(3A\right)=2{\cos }^{3}A-\cos A-2\cos A+2{\cos }^{3}A\\ \therefore \cos \left(3A\right)& =& 4{\cos }^{3}A-3\cos A\end{array}$

Therefore, the required formula is $\cos 3A=4{\cos }^{3}A-3\cos A$.