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Mathematics 20 Online
OpenStudy (madgirlwithabluebox):

Simplify the trigonometric expression. (will medal)

OpenStudy (madgirlwithabluebox):

\[ \frac{ \sin^2\theta}{ 1+\cos \theta}\]

OpenStudy (freckles):

sin^2(theta) can be written as 1-cos^2(theta) and guess what that can be factored

OpenStudy (mathstudent55):

Use the identity below (solved for \(\sin^2 \theta\)) \(\sin^2 \theta + \cos^2 \theta = 1\) and do a substitution in the numerator. Then factor the numerator and reduce.

OpenStudy (madgirlwithabluebox):

Im still really confused

OpenStudy (anonymous):

=(1-cosθ^2)/(1+cosθ)

OpenStudy (anonymous):

=(1-cosθ)(1+cosθ)/(1+cosθ) then we simplfy we know that a^2 - b^2=(a-b)(a+b) @Madgirlwithabluebox

OpenStudy (anonymous):

so we get after simplifying the result =(1-cosθ) did u understand @Madgirlwithabluebox

OpenStudy (madgirlwithabluebox):

Uh no? I get that since there was (1+cosθ)/(1+cosθ) you simplified and got the answer but i dont understand anything else

OpenStudy (anonymous):

so? You understood it or no now ?

OpenStudy (madgirlwithabluebox):

I think ,

OpenStudy (freckles):

does this make sense: assuming y not -1 simplifying (1-y^2)/(1+y) \[\frac{1-y^2}{1+y}=\frac{(1-y)(1+y)}{1+y}=\frac{\cancel{(1+y)}(1-y)}{\cancel{1+y}}=1-y\]

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