# Math Help - Exact value of sin 17pi/12

1. ## Exact value of sin 17pi/12

12.a) Determine an exact value for sin 17pi/12.

b) Determine the answer to part a) using a different method.

So as of now I only know one way to do it.. which is using the compound angle formula. This is what I did:

sin(x+y) = sinxcosy+cosxsiny
sin(2pi/3 + 3pi/4) = sin(2pi/3)cos(3pi/4)+cos(2pi/3)sin(3pi/4)
=(sqr3/2)(-1/sqr2) + (-1/2)(1/sqr2)
=-sqr3/2sqr2 + -1/2sqr2
=-sqr3-1/2sqr2

This is the correct answer. Any ideas on what a second way of finding the exact value of sin 17pi/12 would be? The easier the better haha.

2. Originally Posted by kmjt
12.a) Determine an exact value for sin 17pi/12.

b) Determine the answer to part a) using a different method.

So as of now I only know one way to do it.. which is using the compound angle formula. This is what I did:

sin(x+y) = sinxcosy+cosxsiny
sin(2pi/3 + 3pi/4) = sin(2pi/3)cos(3pi/4)+cos(2pi/3)sin(3pi/4)
=(sqr3/2)(-1/sqr2) + (-1/2)(1/sqr2)
=-sqr3/2sqr2 + -1/2sqr2
=-sqr3-1/2sqr2

This is the correct answer. Any ideas on what a second way of finding the exact value of sin 17pi/12 would be? The easier the better haha.

Code:
>sin(17*pi/12)
-0.965926
>
>(sqrt(3)/2)*(-1/sqrt(2))+(-1/2)*(1/sqrt(2))
-0.965926
>
Your poor use of brackets makes it difficult to tell if you have simplified correctly.

For a second method use:

$\sin(17 \pi/12)=\sin(\pi +5\pi/12)=-\sin(5\pi/12)$

Now proceed as before.

CB

3. Hello, kmjt!

12 (a) Determine an exact value for: . $\sin\tfrac{17\pi}{12}$

. . (b) Determine the answer to part (a) using a different method.

Use the identity: . $\cos2\theta \:=\:1-2\sin^2\!\theta \quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{1-\cos2\theta}{2}}$

Since $\tfrac{17\pi}{12}$ is in Quadrant 3, $\sin\tfrac{17\pi}{12}$ is negative.

We have: . $\sin\frac{17\pi}{12} \;=\;-\sqrt{\frac{1-\cos\frac{17\pi}{6}}{2}} \;=\;-\sqrt{\frac{1-\cos\frac{5\pi}{6}}{2}} \;=\;-\sqrt{\frac{1-(\text{-}\frac{\sqrt{3}}{2})}{2}}$

. . . . . . . . $=\;-\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}} \;=\;-\sqrt{\frac{2+\sqrt{3}}{4}} \;=\; -\frac{\sqrt{2+\sqrt{3}}}{2}$

4. I'm not quite sure whats going on there How is the stuff changing in the square root? Like at first the top is 1 - cos 17pi6 then in the next part its cos 5pi/6?