Why is the trigonometric equation Sin(Ө) = Cos(90 – Ө) an identity?

Question

I need to explain the identities' truth by referring to the unit circle and the definitions of the various trigonometric functions.

Answer 1

draw a radius in first quadrant and note that from perspective of @ the opposite side is y, the adjacent side is x and the hypotenuse is r

now consider the complementary angle 90-@. From its perspective the adjacents side is y and the opposite side is x

sin@ = O/H = y/r = A'/H = cos (90-@)

Answer 2

Because trigonometry is based on right triangle relationships.

The sine of an angle is a right triangle is defined to be the ratio of the opposite leg to the hypotenuse. The cosine is defined to be the adjacent leg over the hypotenuse.

If you subtract Ó¨ from 90 you get the value of the other acute angle in the right triangle.. which means the adjacent leg for this angle (90 - Ó¨) was the opposite leg for Ó¨...

Hence, sin Ó¨ = cos (90 - Ó¨)

Answer 3

Since sin = opposite/hypotenuse and
cosine = adjacent/hypotenuse
it's pretty easy to see that, in a right triangle, if one angle is Φ, then the other angle must be 0-Φ since the 3'rd angle is 90 and the sum of the interior angles of a triangle is 180. Try drawing it on a piece of paper and you'll see it.

HTH

Doug

Answer 4

sin(t) = cos(90 - t) is an identity.

We can use the cosine addition/subtraction identity to verify this.
cos(x - y) = cos(x)cos(y) + sin(x)sin(y)

RHS = cos(90 - t)
RHS = cos(90)cos(t) + sin(90)sin(t)
RHS = (0)cos(t) + (1)sin(t)
RHS = 0 + sin(t) = sin(t) = LHS