Trig: addition formula question...
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Trig: addition formula question...

[From: ] [author: ] [Date: 12-10-31] [Hit: ]
......
how is sin(x-3π/2)=cosx?

could you please explain?

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for any angles, a & b, you have:
sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

substitute x for z and 3π/2 for b

sin(x-3π/2) = sin(x)cos(3π/2) - cos(x)sin(3π/2)

But cos(3π/2) = 0 and sin(3π/2) = -1 Therefore:

sin(x - 3π/2) = sin(x)*0 - cos(x)*(-1) = 0 + cos(x) = cos(x)
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