how to prove this cofuction identity: sin x = cos (pi/2-x)?

Question

help me please.
is it true for any value of x?
thank u!

Answer 1

sin(x) = cos(pi/2 - x)

To prove this, all you have to do is use the cosine addition identity, which states that
cos(a - b) = cos(a)cos(b) + sin(a)sin(b).

RHS = cos(pi/2 - x)
= cos(pi/2)cos(x) + sin(pi/2)sin(x)
= (0)cos(x) + (1)sin(x)
= sin(x)
= LHS

Answer 2

Since Cos(-x) = cos(x) You can show:

Cos(pi/2 - x) = cos(x-pi/2) (multiply the interior by -1)

Since sin(x) and cos(x) are offset by pi/2, the phase shift in your cos(x-pi/2) will equate the two.

Yes, it is true for ANY value of x

Answer 3

cos(a-b) = cos(a)*cos(b) + sin(a)*sin(b)
a = pi/2
b = x

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

Answer 4

cos((pi/2) - x) =
cos(pi/2)cos(x) + sin(pi/2)sin(x) =
0cos(x) + 1sin(x) =
sin(x)

so

cos((pi/2) - x) = sin(x)