Question

What are the sine cosine and tangent of theta=7pi/4 radians

Answer 1

Answer:

see explanation

Step-by-step explanation:

$\frac{7\pi }{4}$ is in the fourth quadrant

Where sin and tan are < 0 , cos > 0

The related acute angle is 2π - $\frac{7\pi }{4}$ = $\frac{\pi }{4}$

Hence

sin([$\frac{7\pi }{4}$ ) = - sin($\frac{\pi }{4}$) = - $\frac{1}{\sqrt{2} }$ = - $\frac{\sqrt{2} }{2}$

cos($\frac{7\pi }{4}$) = cos($\frac{\pi }{4}$) =  $\frac{\sqrt{2} }{2}$

tan($\frac{7\pi }{4}$= - tan($\frac{\pi }{4}$ = - 1

Answer 2

Answer:

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

$\cos\theta=\frac{1}{\sqrt{2}}$

$\tan\theta=-1$.

Step-by-step explanation:

We have to find the value of sine cosine and tangent of $\theta=\frac{7\pi}{4}$ radians.

$\frac{7\pi}{4}\times \frac{180}{\pi}=315^{\circ}$

So, $\theta=\frac{7\pi}{4}$ lies in 4th quadrant. Sine and tangent are negative in 4th quadrant.

The value of sinθ is

$\sin\theta=\sin (\frac{7\pi}{4})$

$\sin\theta=\sin (2\pi-\frac{\pi}{4})$

$\sin\theta=-\sin (\frac{\pi}{4})$

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

The value of cosθ is

$\cos\theta=\cos (\frac{7\pi}{4})$

$\cos\theta=\cos (2\pi-\frac{\pi}{4})$

$\cos\theta=\cos (\frac{\pi}{4})$

$\cos\theta=\frac{1}{\sqrt{2}}$

The value of tanθ is

$\tan\theta=\tan (\frac{7\pi}{4})$

$\tan\theta=\tan (2\pi-\frac{\pi}{4})$

$\tan\theta=-\tan (\frac{\pi}{4})$

$\tan\theta=-1$

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