(sin130)^2 + (cos50)^2

Let us rewrite:

= (sin(180-50))^2 + (cos50)^2

We know that: sina = sin (180-a)

(the sine is positive in 1st and 2nd quadrants )

==> (sin(180-50))^2 + (cos50)^2= (sin50)^2 + (cos50)^2 = 1

we know that: (cosa)^2 + (sina)^2 = 1

==> **(sin130)^2 + (cos50)^2 = 1**

To show (sin130)^2 + (cos50)^2

Solution:

Sin 130 = sin(180-50) = sin50.

Therefore (sin 50)^2+cos(50)^2 = 1, as ( sinx)^2+(cosx)^2 = 1 is a trigonometric identity.

So (sin150)^2+(cos50)^2 = 1

It is obvious that sin 130 = sin 50.

We'll prove that using the formula:

sin (a-b) = sin a * cos b - sin b* cos a

We'll put a = 180 and b = 130

sin(180-50) = sin 130 = sin 180*cos 50 - sin 50 * cos 180

But sin 180 = 0 and cos 180 = -1

sin(180-50) = 0 - (- sin 50)

sin 130 = sin 50

We'll substitute sin 130 by sin 50 and we'll get:

**(sin 50)^2 + (cos 50)^2 = 1** (fundamental formula of trigonometry)