Question

How do you find the values of $\sin 2\theta$ and $\cos 2\theta$ when $\cos \theta =\frac{12}{13}$?

Answer 1

Step 1: Calculate the perpendicular side

We are given that $\cos \theta =\frac{12}{13}$

We know that

$\cos \theta =\frac{b}{h}=\frac{12}{13}$

where $b$ is the base of a triangle, and $h$ is the hypotenuse of the triangle.

Let us consider

$b=12$ and $h=13$

According to Pythagoras theorem

$h^2=p^2+b^2$

where $p$ is the perpendicular of a triangle.

$\therefore p^2=h^2-b^2$

$={13}^2-{12}^2$

$=169-144$

$=25$

$\therefore p=\sqrt{25}$

$=5$ units

Step 2: Calculate $\sin 2\theta$

Therefore,

$\sin \theta =\frac{5}{13}$

$\cos \theta =\frac{12}{13}$

We know that,

$\sin 2\theta =2\sin \theta \cos \theta$

$=2\times \frac{5}{13}\times \frac{12}{13}$

$=\frac{120}{169}$

Step 3: Calculate $\cos 2\theta$

We know that,

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

$={\left(\frac{12}{13}\right)}^2-{\left(\frac{5}{13}\right)}^2$

$=\frac{144-25}{169}$

$=\frac{119}{169}$

Hence, the required values are $\frac{120}{169}$ and $\frac{119}{169}$.