Question

If $\theta ={30}^o$, prove that $4{\cos }^2\theta -3\cos 0^0=\cos 3\theta$

Answer 1

In the given identity $4{\cos }^2\theta -3\cos 0^0=\cos 3\theta$, substitute $\theta ={30}^0$ as shown below:

$4{\cos }^2{30}^0-3\cos 0^0=\cos \left({3\times 30}^0\right)\Rightarrow 4{\cos }^2{30}^0-3\cos 0^0=\cos {90}^0$

We know that the values of the trignometric functions $\cos {30}^0=\frac{\sqrt{3}}{2},$ $\cos 0^0=1$ and$\cos {90}^0=0$.

Let usfirstfind the value of $4{\cos }^2{30}^0-3\cos 0^0$ as shown below:

$4{\cos }^2{30}^0-3\cos 0^0=1-\left({4\times \left(\frac{\sqrt{3}}{2}\right)}^2\right)-\left(3\times 1\right)=\left(4\times \frac{3}{4}\right)-3=3-3=0...........\left(1\right)$

We know that $\cos {90}^0=0...........\left(2\right)$

From equations 1 and 2, we get that $4{\cos }^2{30}^0-3\cos 0^0=\cos {90}^0$.

Hence, $4{\cos }^2\theta -3\cos 0^0=\cos 3\theta$ when $\theta ={30}^0$.