Trig maths question?

Question

sin165 as sin(120+45) show that sin165 = root6-root2 / 2

I do get the answer but I get a plus sign not a negative...

Please help

Answer 1

This follows the theorem sin(α+ß)=sin α·cos ß + cos α·sin ß
So, α=120 and ß=45.

sin(120+45) = sin(120)·cos(45) + cos(120)·sin(45)
= (√3/2)(√2/2)+(-1/2)(√2/2)
=√6/4 - √2/4
=(√6 - √2)/4

1) You forgot that cos(120) is -1/2, not +1/2
2) For your denominators, I think you used √2 instead of regular ol' 2. So your final answer's denominator should be 4, not 2.

Answer 2

Assuming that al measures are in degrees....

sin(120+45)=sin(120)cos(45)
+cos(120)sin(45)

=(sqrt(3)/2)(sqrt(2)/2)-(1/2)(sqrt(2)/2)
=(sqrt(6)-sqrt(2))/4

Notes:

1. Denominator should be 4, not 2.
2. You missed the fact that 120 degrees is in quadrant II, so cosine is negative.

Answer 3

sin(x+y) = sin(x)*cos(y)+sin(y)*cos(x)
Sin(120+45) = sin(120)*cos(45) + sin(45)*cos(120)

sin(120) =sqrt(3)/2
sin(45) = cos(45) = sqrt(2)/2
cos(120)= -1/2

SO sin(120+45) = sqrt(6)/4 - sqrt(2)/4