Question

please help?
Use the product-to-sum formula to write the given product as a sum or difference

12sin(pi/6)sin(pi/6)

Answer 1

Do you have a specific product-to-sum formula? Or is it the one I'm thinking of?
\[\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y\\
\sin(x\pm y)=\sin x\cos y\pm \sin y\cos x\]

Answer 2

it doesnt day a specific one

Answer 3

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmth/trid2.gif

Answer 4

I gather your case there uses the last indentity listed in there

Answer 5

ANSWER CHOICES:
6sin(pi/12)
6-6cos(pi/3)
6+6cos(pi/12)
-6sin(pi/12)
6sin(pi/6)+6cos(pi/6)

if that helps

Answer 6

so... what do you get using the trig identity there?

Answer 7

I think its 6+6cos(pi/12)?

Answer 8

I got something else :/

Answer 9

you walk me through the steps? i did it using that and got a few different answers

Answer 10

\(\bf 12\left[sin \left(\cfrac{\pi}{6}\right)sin \left(\cfrac{\pi}{6}\right)\right]\\
\left[sin \left(\cfrac{\pi}{6}\right)sin \left(\cfrac{\pi}{6}\right)\right] =
\frac{1}{2}\left[ cos\left(\cfrac{\pi}{6}-\cfrac{\pi}{6}\right)-cos\left(\cfrac{\pi}{6}+\cfrac{\pi}{6}\right)\right]\)

Answer 11

so notice the 1st pair subtracting from each other, what would \(\bf cos\left(\cfrac{\pi}{6}-\cfrac{\pi}{6}\right)\) be equal to?

Answer 12

cos0?

Answer 13

yes
what about the 2nd pair, what would \(\bf cos\left(\cfrac{\pi}{6}+\cfrac{\pi}{6}\right)\)
be equals to?

Answer 14

sorry im confused.. im going guess
12(pi/6)

Answer 15

well, let's leave the "12" on the side for now, is just a multiplier of the identity anyway

Answer 16

1/6 + 1/6 = ?

Answer 17

1/3

Answer 18

thus \(\bf \left[sin \left(\cfrac{\pi}{6}\right)sin \left(\cfrac{\pi}{6}\right)\right] =
\frac{1}{2}\left[ cos\left(0\right)-cos\left(\cfrac{2\pi}{6}\right)\right]\\ \implies
\frac{1}{2}\left[ cos\left(0\right)-cos\left(\cfrac{\pi}{3}\right)\right]\\
\)

Answer 19

so, what would be the cos(0)?

Answer 20

pi/3 or 1/3?

Answer 21

well, check your unit circle for the cos(0), or even your calculator :)

Answer 22

oh its 1

Answer 23

\(\bf \frac{1}{2}\left[ cos\left(0\right)-cos\left(\cfrac{\pi}{3}\right)\right] \implies
\frac{1}{2}\left[ 1-cos\left(\cfrac{\pi}{3}\right)\right]\\
\cfrac{12\left[1-cos\left(\cfrac{\pi}{3}\right)\right]}{2}\)

Answer 24

so now we bring in the multiplier, 12

Answer 25

as you can pretty much see what the answer is :)

Answer 26

\(\bf \cfrac{12\left[1-cos\left(\cfrac{\pi}{3}\right)\right]}{2}\\
\cfrac{\cancel{12}-\cancel{12}cos\left(\cfrac{\pi}{3}\right)}{\cancel{2}} \implies
6-6cos\left(\cfrac{\pi}{3}\right)\)

Answer 27

ok thank you! i was working on it haha

Answer 28

hehe

Answer 29

yw