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

Question

asked Feb 1, 2021 111k views

2 Answers

Vijay Sahu · Answer 1 · 2021-02-06T02:01:15+0000

Answer:

$\sin\theta=-(\sqrt2)/(2)}$ and
$\tan\theta=-1$

Explanation:

Given :
$\cos\theta=(\sqrt2)/(2)$ and
$(3\pi)/(2)<\theta<2$

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

Solution :

As
$(3\pi)/(2)<\theta<2$ 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,

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

$\sin\theta=√(1-\cos^2\theta)$

Substitute
$\cos\theta=(\sqrt2)/(2)$ ,

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

$\sin\theta=\pm\sqrt{1-(2)/(4)}$

$\sin\theta=\pm\sqrt{1-(1)/(2)}$

$\sin\theta=\pm\sqrt{(1)/(2)}$

$\sin\theta=-(\sqrt2)/(2)}$

Now using another formula,

$\tan\theta=(\sin\theta)/(\cos\theta)$

$\tan\theta=(-(\sqrt2)/(2))/((\sqrt2)/(2))$

$\tan\theta=-1$

Therefore,
$\sin\theta=-(\sqrt2)/(2)}$ and
$\tan\theta=-1$ .