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Trigonometry 20 Online
OpenStudy (anonymous):

(1-cos^2x)(cscx)

OpenStudy (tkhunny):

You do know that \(\sin^{2}(x) = 1 - \cos^{2}(x)\)?

OpenStudy (anonymous):

no

OpenStudy (tkhunny):

Are you sure? Maybe \(\sin^{2}(x) + \cos^{2}(x) = 1\) looks more familiar?

OpenStudy (anonymous):

is that what you think the answer is?

OpenStudy (tkhunny):

It's called a hint.

OpenStudy (anonymous):

well can you explain step by step how to so it.

OpenStudy (anonymous):

*do

OpenStudy (tkhunny):

I did, already, in my first post. Take a good, hard look at that and ponder your original expression.

OpenStudy (anonymous):

try replacing cscx with its reciprocal.... then multiply....

OpenStudy (anonymous):

answer is\[\sin x\] You know csc x = 1/sinx, and cos^2x+sin^2x = 1. So 1-cos^2x = sin^x, and sin^2x*cscx = sin x

OpenStudy (abb0t):

NOTE: \[\csc(x) = \frac{ 1 }{ \sin(x) }\] Similarly, \[\cot(x) = \frac{ \cos(x) }{ \sin(x) }\] \[\tan(x) = \frac{ \sin(x) }{ \cos(x) }\] \[\sec(x) = \frac{ \ 1 }{ \cos(x) }\] \[1-\cos^2(x) = \sin^2(x) \]

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