Trigonomatric Identities

**soni** · Aug 10th 2010, 11:40 PM

a. Find the exact value of sin 180° by using sin (270 °– 90 °)

b. Find the exact value of sin 150° by using sin (210 °– 60 °)

c. Use the value of sin 150° found in b to find sin (–15 °) by using sin
(150 °– 165 °)

**Vlasev** · Aug 11th 2010, 12:09 AM

For part a) you need to use the summation formula for the sine function. You have to recognize that sin(270) = -Sin(90) = -1 and cos(270) = cos(90) = 0.

Then you have
$<br /> \sin(270-90) = \sin(270)\cos(90) - \sin(90)\cos(270) = (-1)(0) - (1)(0) = 0$

This should not be surprising since 270-90 = 180 and we know that the sine function is zero there. It is a very similar situation in b) and c) and you should be able to do it yourself. Try it out and if you can't do it, show your work and ask for help!

**masters** · Aug 11th 2010, 06:36 AM

Originally Posted by soni

b. Find the exact value of sin 150° by using sin (210 °– 60 °)

Hi soni,

$\sin(u \pm v)=\sin u \cos v \pm \cos u \sin v$

$\sin(210-60)=\sin 210 \cos 60-\cos 210 \sin 60=\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)-\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\ right)=.5$

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