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Old 08-07-2007   #1
tot_6nxs
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Default A specific Trigonometry question....Thanks in advance?

Find the exact numbers a and b which make the following identity true:sin(pi/2-x)=acosx+bsinxA brief description of answer and process would be great. Thanks guys/gals!
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Old 08-07-2007   #2
Edgar Greenberg
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sin(X +Y) = sin X cos Y + cos X sin Ysin (π/2 - x) = sin π/2 cos(-x) + cos(π/2)sin(-x)Now sin π/2 = 1, cos(π/2) =0, cos(-x) = cos (x), and sin(-x) = -sin(x). So sin(π/2 - x) = 1cos (x) + -0sin(x) = 1*cos x + 0*sin x.a = 1, b = 0.EDIT: The guy below me is right. In this case, there will be other specific answers, depending on x. For example if x = 0, sin(π/2) = 1, and a could be any number.
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Old 08-07-2007   #3
firat c
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Start with sin(pi/2-x) = cos x.Rewrite to getcos x = a cos x + b sin x,reorganize(1-a) cos x = b sin x.It is easy to see that a=1 and b=0 will give a solution for any x. But for a particular x, there are infinitely many solutions. Assuming x is not zero and a is not 1, divide both sides by sinx and 1-a to getcos x / sin x =b/(1-a).So you just need tan x and two numbers whose quotient is tan x. There are infinitely many possibilities for that and you need to especially take care of the situations when x=0.
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