Question

HELP!!! Determine the exact value of sin 13pi/12

Answer 1

13pi/12 = (1/2)(13pi/6)

Answer 2

pi/12 = pi/3-pi/4

Answer 3

answer in the textbook is... \[\frac{ \sqrt{2}-\sqrt{6} }{ 4 }\] btw

Answer 4

That is not important. Please use one or both of the hints given and produce the result.

Answer 5

The thing is I have no idea how to produce the result... lol

Answer 6

If that is true, you should not be in this class and you will fail. Fortunately, it is not true.

I do not believe you have seen neither a double angle formula for sine nor the expansion formula for the sine of a difference. Please show these identities:

\(\sin(2x) = ??\)

\(\sin(A-B) = ??\)

Answer 7

Listen to these folks. You will learn. This will be easy once you see the solution.

Answer 8

\[\sin (x-y)=\sin x \cos y-\cos x \sin y\]

Answer 9

yeah i've seen them, its just I don't know how to use them..

\[\sin (A-B) = SinACosB -CosASinB\]
\[Sin(2x)=2sinxcox\]

I'm just having trouble using them

Answer 10

Replace x with pi/3
Replace y with pi/4

Answer 11

"yeah i've seen them," <== This is evidence that you have more than "no idea". Good work.

Answer 12

I'm just asking for help... because I don't know how to find the exact value for this question -.-

Answer 13

\[\sin (A-B)=\sin A \cos B-\cos A \sin B\]
\[\sin (\frac{\pi}{3}-\frac{\pi}{4})=\sin \frac{\pi}{3}\cos \frac{\pi}{4}-\cos \frac{\pi}{3}\sin \frac{\pi}{4}\]
\[\sin \frac{\pi}{12}=\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}-\frac{1}{2}\frac{\sqrt{2}}{2}\]
\[\sin \frac{\pi}{12}=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}\]

Answer 14

\(\sin(2(13\pi/6)) = 2\sin(13\pi/6)\cos(13\pi/6)\)

You should be able to determine the exact values of the two functions on the right hand side. 6ths are standard values.

Answer 15

Now, since 13pi/12 is in quadrant III, the sin will be negative so change the answer for sin pi/12 to its opposite.

Answer 16

oh okay, thanks so much for you help!!

Answer 17

yw