Question 550210: About trigonometric functions?
Given sample expression:
sinθ+sin(θ+π/2)+sin(θ+π)+sin(θ+3/2π)
= sinθ+cosθ-sinθ+sin[(θ+π)+π/2]
= cosθ+cos(θ+π)
= cosθ-cosθ
my problem is that as i tried to figure this out.. i got confused in this part
"sin[(θ+π)+π/2]" <----- how did it arrived to "cos(θ+π)"
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! my problem is that as i tried to figure this out.. i got confused in this part
"sin[(θ+π)+π/2]" <----- how did it arrived to "cos(θ+π)"
**
Think of rotating reference angle θ in quadrant I, π radians which places the reference angle in quadrant III. Then rotate angle (θ+π) another π/2 radians which places the reference angle in quadrant IV. But this last reference angle is the compliment of (θ+π), thus it converts from sin to cos.
..
This can be made clearer by an example as follows:
let θ=30º in quadrant I, reference angle=30º
30+π=210ºin quadrant III, reference angle=30º
210+π/2=300º in quadrant IV, reference angle=60º
So,sin[(30+π)+π/2]=cos(30+π)
hope this helps.
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