HELP PLEZ TRIGONOMETRY!

Question

asked Mar 19, 2021 115k views

1 Answer

Scott Rice · Answer 1 · 2021-03-25T07:15:12+0000

$(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha$

Solution:

Given that we have to simplify:

$(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha)$ ---- eqn 1

We know that,

sin^2 x = 1 - cos^2 x

Substitute the above identity in eqn 1

$(2\left(1-\cos ^(2) \alpha\right)-1)/(\sin \alpha+\cos \alpha)$

Simplify the above expression

$(2-2 \cos ^(2) \alpha-1)/(\sin \alpha+\cos \alpha)$

$(1-2 \cos ^(2) \alpha)/(\sin \alpha+\cos \alpha)$ ------- eqn 2

By the trignometric identity,

(sin x + cos x)(sin x - cos x) = 1-2cos^2 x

Substitute the above identity in eqn 2

$((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)$

Cancel the common factors in numerator and denominator

$((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)=\sin \alpha-\cos \alpha$

Thus the simplified expression is:

$(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha$