A specific Trigonometry question....Thanks in advance? - XP Math - Forums

 XP Math - Forums A specific Trigonometry question....Thanks in advance?

 08-07-2007 #1 tot_6nxs Guest   Posts: n/a 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!
 08-07-2007 #2 Edgar Greenberg Guest   Posts: n/a sin(X +Y) = sin X cos Y + cos X sin Ysin (Ï€/2 - x) = sin Ï€/2 cos(-x) + cos(Ï€/2)sin(-x)Now sin Ï€/2 = 1, cos(Ï€/2) =0, cos(-x) = cos (x), and sin(-x) = -sin(x). So sin(Ï€/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(Ï€/2) = 1, and a could be any number.
 08-07-2007 #3 firat c Guest   Posts: n/a 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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