# Equivalent expression for sin(-7pi/6) in terms of the related acute angle.

• Nov 27th 2009, 06:31 AM
kmjt
Equivalent expression for sin(-7pi/6) in terms of the related acute angle.
State an equivalent expression for sin (-7pi/6) in terms of the related acute angle.

I don't really know how to solve these kind of questions, I find everything I google doesn't explain how to do it :(
• Nov 27th 2009, 06:59 AM
Quote:

Originally Posted by kmjt
State an equivalent expression for sin (-7pi/6) in terms of the related acute angle.

I don't really know how to solve these kind of questions, I find everything I google doesn't explain how to do it :(

HI

Put it in degrees so that its easier for you to see.

$\displaystyle \sin (-\frac{7\pi}{6})=\sin (-210)=-\sin 210=\sin 30=0.5$

Note that $\displaystyle \sin 30=\sin 150=-\sin 210=-sin 330$
• Nov 27th 2009, 07:25 AM
kmjt
I'm not following =/ I know that sin (-7pi/6) is the same thing as sin (-210 degrees) but i'm not sure where to go from there. I'm not even sure what the question is asking (Worried)
• Nov 27th 2009, 07:34 AM
Quote:

Originally Posted by kmjt
I'm not following =/ I know that sin (-7pi/6) is the same thing as sin (-210 degrees) but i'm not sure where to go from there. I'm not even sure what the question is asking (Worried)

OK , try to follow here . I will go slow .

$\displaystyle \sin (-210)=-\sin 210$ (Step 1)

$\displaystyle -\sin 210=-(-\sin 30)$ (Step 2)

This is because sin is negative in the 3rd quadrant .

$\displaystyle -(-\sin 30)=\sin 30$ (Step 3)

I think this is what the question wants since it asks for acute angles (angle < than 90 degree)

If you still have any problem , just indicate which step you are unsure with .. don say everything or else i don know where to start .
• Nov 27th 2009, 08:01 AM
kmjt
Where is the-(-sin30) coming from?
• Nov 27th 2009, 08:05 AM
Quote:

Originally Posted by kmjt
Where is the-(-sin30) coming from?

sin 210 = - sin 30

- sin 210 = - ( - sin 30 )
• Nov 27th 2009, 08:12 AM
kmjt
It just clicked in my brain thanks!
• Nov 27th 2009, 06:35 PM
HallsofIvy
I don't see why degrees should be easier than radians and I think it is good to encourage people to think in terms of radians. $\displaystyle \frac{7\pi}{6}= \pi+ \frac{\pi}{6}$. Since the angle is negative, we swing down from the positive x axis to the negative y axis and then $\pi/6$ above. Now we are in the second quadrant where sine is still postive.

$\displaystyle sin(-\frac{7\pi}{6})= sin(\frac{\pi}{6})$.