Question

Prove that $2{\cos }^2{30}^o-1=1-2{\sin }^2{30}^o=\cos {60}^o$

Answer 1

We know that the values of the trignometric functions $\cos {30}^0=\frac{\sqrt{3}}{2},$ $\sin {30}^0=\frac{1}{2}$ and$\cos {60}^0=\frac{1}{2}$.

Let usfirstfind the value of $2{\cos }^2{30}^0-1$ as shown below:

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

$=\left(2\times \frac{3}{4}\right)-1$

$=\frac{3}{2}-1$

$=\frac{3-2}{2}$

$=\frac{1}{2}...........\left(1\right)$

Now, finding the value of $1-2{\sin }^2{30}^0$ as follows:

$1-2{\sin }^2{30}^0=1-\left({2\times \left(\frac{1}{2}\right)}^2\right)$

$=1-\left(2\times \frac{1}{4}\right)$

$=1-\frac{1}{2}$

$=\frac{2-1}{2}$

$=\frac{1}{2}...........\left(2\right)$

We also know that,

$\cos {60}^0=\frac{1}{2}...........\left(3\right)$

From equations $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get:

$2{\cos }^2{30}^0-1=1-2{\sin }^2{30}^0=\cos {60}^0=\frac{1}{2}$.

Hence, it is proved that $2{\cos }^2{30}^0-1=1-2{\sin }^2{30}^0=\cos {60}^0$.