A specific Trigonometry question....Thanks in advance?

Question

Find the exact numbers a and b which make the following identity true:

sin(pi/2-x)=acosx+bsinx

A brief description of answer and process would be great. Thanks guys/gals!

Answer 1

sin(X +Y) = sin X cos Y + cos X sin Y
sin (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.

Answer 2

Start with
sin(pi/2-x) = cos x.
Rewrite to get
cos 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 get
cos 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.

Answer 3

since sin(π/2 - x) = cos x, you have
cos x = a cos x + b sin x,
which makes a = 1 and b = 0.