Question

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

Answer 1

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`