How do I verify this equation is an identity? sin(180°-α)=1-cos^2α/sinα

1 Answer
Mar 18, 2018

Verified below

Explanation:

Here's two identities you will need:
sin(theta-α)= sintheta*cosα-costheta*sinα
sin^2α+cos^2α=1

Start:
sin(180°-α)=(1-cos^2α)/sinα

sin(180°)*cosα-cos(180°)*sinα=(1-cos^2α)/sinα

cancel((0)*cosα)-(-1)*sinα=(1-cos^2α)/sinα

sinα=(1-cos^2α)/sinα

sin^2α/sinα=(1-cos^2α)/sinα

(sin^2α+1-1)/sinα=(1-cos^2α)/sinα

Substitute in: sin^2α+cos^2α for 1:

(sin^2α+1-(sin^2α+cos^2α))/sinα=(1-cos^2α)/sinα

((cancel(sin^2α)+1cancel(-sin^2α)-cos^2α))/sinα=(1-cos^2α)/sinα

(1-cos^2α)/sinα=(1-cos^2α)/sinα