SOLUTION: 2cos(x-π/2)+3sin(x+π/2)-(3sinx+2cosx)=?

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Question 1044442: 2cos(x-π/2)+3sin(x+π/2)-(3sinx+2cosx)=?
Found 2 solutions by Cromlix, ikleyn:
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
2cos(x-π/2)+3sin(x+π/2)-(3sinx+2cosx)=
Multiply out.
2cos(x) - 2cos(π/2) + 3sin(x) + 3sin(π/2) - 3sin(x) - 2cos(x)
cosπ/2 = 0
sinπ/2 = 1
So equation is now:-
2cos(x) + 3sin(x) + 3 - 3sin(x) - 2cos(x)
(2cos(x)'s cancel out)
(3sin(x)'s cancel out)
Answer
= 3.
Hope this helps :-)

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
2cos(x-pi/2)+3sin(x+pi/2)-(3sinx+2cosx)=?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

1.  cos%28x+-+pi%2F2%29 = cos%28x%29%2Acos%28pi%2F2%29+%2B+sin%28x%29%2Asin%28pi%2F2%29 = cos(x)*0 + sin(x)*1 = sin(x).

2.  sin%28x%2Bpi%2F2%29 = sin%28x%29%2Acos%28pi%2F2%29+%2B+cos%28x%29%2Asin%28pi%2F2%29 = sin(x)*0 + cos(x)*1 = cos(x).

3.  Therefore

    2cos%28x-pi%2F2%29%2B3sin%28x%2Bpi%2F2%29-%283sinx%2B2cosx%29 = 2sin(x) + 3cos(x) -(3sin(x) + 2cos(x)) = -sin(x) + cos(x).

Answer.  2cos%28x-pi%2F2%29%2B3sin%28x%2Bpi%2F2%29-%283sinx%2B2cosx%29 = cos(x) - sin(x).