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

Answer 1

Brooklyn Anderson

Verified below

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#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.