Question

Cos (0) = square root 2/2 and 3pi/2<0<2 pi , evaluate sin(0) and tan (0)

Answer 1

Answer:

$\sin\theta=-\frac{\sqrt2}{2}}$ and $\tan\theta=-1$

Step-by-step explanation:

Given : $\cos\theta=\frac{\sqrt2}{2}$ and $\frac{3\pi}{2}$

To find : Evaluate $\sin\theta$ and $\tan\theta$ ?

Solution :

As $\frac{3\pi}{2}$ means $\theta$ belong to the third or fourth quadrant .

So,   $\sin\theta$ is negative and $\tan\theta$ is positive in third and negative in fourth.

Using the formula of trigonometric,

$\sin^2\theta+\cos^2\theta=1$

$\sin\theta=\sqrt{1-\cos^2\theta}$

Substitute  $\cos\theta=\frac{\sqrt2}{2}$ ,

$\sin\theta=\pm\sqrt{1-(\frac{\sqrt2}{2})^2}$

$\sin\theta=\pm\sqrt{1-\frac{2}{4}}$

$\sin\theta=\pm\sqrt{1-\frac{1}{2}}$

$\sin\theta=\pm\sqrt{\frac{1}{2}}$

$\sin\theta=-\frac{\sqrt2}{2}}$

Now using another formula,

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

$\tan\theta=\frac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}$

$\tan\theta=-1$

Therefore, $\sin\theta=-\frac{\sqrt2}{2}}$ and $\tan\theta=-1$.