trig ratios

• May 3rd 2010, 05:47 AM
sigma1
trig ratios
write each of the following as the ratios of positive acute angles.

in the examples i understand that $sin 330 = -sin 30$

but can someone please explain to me how it is that

$cos (3/4) \pi$= $-cos (1/4) \pi$

and $sin -(4/5) \pi$= $-sin (1/5) \pi$

i understand that $\pi$ = 180 but i am not geting it when i express each of the sectors in terms of $\pi$
• May 3rd 2010, 06:24 AM
masters
Quote:

Originally Posted by sigma1
write each of the following as the ratios of positive acute angles.

in the examples i understand that $sin 330 = -sin 30$

but can someone please explain to me how it is that

$cos (3/4) \pi$= $-cos (1/4) \pi$

and $sin -(4/5) \pi$= $-sin (1/5) \pi$

i understand that $\pi$ = 180 but i am not geting it when i express each of the sectors in terms of $\pi$

Hi sigma1,

$\cos \frac{3 \pi}{4}$ is a QII angle. Related angle is $\pi - \frac{3 \pi}{4}=\frac{\pi}{4}$.
Cosine is negative in the second quadrant.

$\cos \frac{\pi}{4}$ is a QI angle. Cosine is positive in the first quadrant.

Therefore, $\cos \frac{3 \pi}{4}=-\cos \frac{\pi}{4}$

The sin function is an odd function, meaning $\sin (-\theta)=-\sin \theta$

$\sin -\frac{4 \pi}{5}=-\sin \frac{4\pi}{5}$

Related angle is $\pi-\frac{4\pi}{5}=\frac{\pi}{5}$

$\sin \frac{4\pi}{5}$ is a QII angle. Sin is positive in the first and second quadrants.

Therefore, $\sin -\frac{4\pi}{5}=-\sin \frac{\pi}{5}$