Question

I need to figure out how to get the exact value of sin13pi/12

im trying to do addition of sin ((3pi/4)+(pi/3))

but I cant use the addition formula when the angle is outsides of the first quadrant.

im trying to use sin(a+b)= sinAcosB+cosAsinB

but it isnt coming out. can someone explain to me why or how am i supposed to do this problem?

Answer 1

why not?

Answer 2

formula works, doesn't depend on "quadrant"

Answer 3

not to mention that fact that both \(\frac{3\pi}{4}\) and \(\frac{\pi}{3}\) are in quadrant 1, not that it matters

Answer 4

hmm alright here is what I have then

(isnt 3pi/4 in Q2? )

sin3pi/4 cos pi/3 + cos 3pi/4 sin pi/3

sqrt2/2 (1/2) + -sqrt2/2(sqrt3/2)

sqrt2/4 + -(sqrt6/4)

(sqrt 2 -sqrt6 )/ 4

thats how I run it down, but the problem I have is thats the wrong answer

I need to get (sqrt6-sqrt2)/4

also when I try to put
sin (3pi/4) cos(pi/3) +cos(3pi/4)sin(pi/3) into wolframalpha. it says that my input is different from 13pi/12

Answer 5

Thank you for your help!

Answer 6

for example.

http://www.wolframalpha.com/input/?i=sin+%28%283pi%2F4%29%2B%28pi%2F3%29%29

this is the link when I input sin((3pi/4)+(pi/3))


http://www.wolframalpha.com/input/?i=sin+%283pi%2F4%29+cos%28pi%2F3%29+%2Bcos%283pi%2F4%29sin%28pi%2F3%29

sin (3pi/4) cos(pi/3) +cos(3pi/4)sin(pi/3)

they arent the same inputs and I dont know why. im obviously using the formula incorrectly. I kinda assumed because It was to the the quadrants and the signs on each trig function.

Answer 7

im sorry I finally figured it out. lol....

((sqrt2-sqrt6)/4 )= -(-1+sqrt(3))/(2 sqrt(2))

It just looks different