`cos(pi - theta) + sin(pi/2 + theta) = 0` Prove the identity.

You need to use the formula `cos(a-b) = cos a*cos b + sin a*sin b` and `sin(a+b) = sin a*cos b + sin b*cos a,` such that:

`cos (pi - theta) = cos pi*cos theta + sin pi*sin theta`

Since `cos pi = -1` and `sin pi = 0` , yields:

`cos (pi - theta) = - cos theta`

`sin(pi/2 + theta) = sin(pi/2)*cos theta + sin theta*cos(pi/2)`

Since `sin(pi/2) = 1` and `cos(pi/2) = 0,` yields:

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You need to use the formula `cos(a-b) = cos a*cos b + sin a*sin b` and `sin(a+b) = sin a*cos b + sin b*cos a,` such that:

`cos (pi - theta) = cos pi*cos theta + sin pi*sin theta`

Since `cos pi = -1` and `sin pi = 0` , yields:

`cos (pi - theta) = - cos theta`

`sin(pi/2 + theta) = sin(pi/2)*cos theta + sin theta*cos(pi/2)`

Since `sin(pi/2) = 1` and `cos(pi/2) = 0,` yields:

`sin(pi/2 + theta) = cos theta`

Replacing `- cos theta` for `cos (pi - theta)` and `cos theta` for in `(pi/2 + theta)` yields:

`cos (pi - theta) + sin(pi/2 + theta) = - cos theta + cos theta = 0`

Hence, checking if the identity is valid yields that `cos (pi - theta) + sin(pi/2 + theta) = 0` holds.