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Trigonometry 14 Online
OpenStudy (anonymous):

prove the identity is true (1-cos^2x)cotx=sinxcosx

OpenStudy (mathlegend):

Hi, do you have an identity sheet that has Pythagorean identities on it?

OpenStudy (mathlegend):

Can you tell me what \[1-\cos ^{2}x \] equals?

OpenStudy (mathlegend):

@hconejo100

OpenStudy (anonymous):

sin^2x

OpenStudy (mathlegend):

and what does cotx also equal is it... \[\frac{ cosx }{ sinx }\] ?

OpenStudy (anonymous):

yes

OpenStudy (mathlegend):

So lets see what we have...

OpenStudy (mathlegend):

\[\sin ^{2}x(\frac{ cosx }{ sinx })\]

OpenStudy (mathlegend):

which simplifies to... \[\frac{ \sin ^{2}xcosx }{ sinx } = sinxcosx\]

OpenStudy (mathlegend):

Make sense?

OpenStudy (anonymous):

not really

OpenStudy (mathlegend):

What step does not make sense?

OpenStudy (anonymous):

well it does makes sense but on my homework i have to prove it by writing it step by step and what rule is each step

OpenStudy (mathlegend):

\[1 - \cos ^{2}x = \sin ^{2}x\] is substitution

OpenStudy (mathlegend):

\[cotx = \frac{ cosx }{ sinx }\] Substitution

OpenStudy (mathlegend):

\[\sin ^{2}x(\frac{ cosx }{ sinx })\] Multiplication

OpenStudy (mathlegend):

\[\frac{ \sin ^{2}xcosx }{ sinx } = sinxcosx\] Simplifying

OpenStudy (mathlegend):

better @hconejo100 ?

OpenStudy (anonymous):

the rules i can chose from are -algebra -reciprocal -quotient -Pythagorean -odd/even here is an example my teacher gave me cotx/cscx-1=cscx+1/cotx to prove it he put =cotx/cscx-1*cscx+1/cscx+1 Algebra =cotx(cscx+1)/csc^2x-1 Algebra =cotx(cscx+1)/cot^2x Pythagorean =cscx+1/cotx algebra

OpenStudy (mathlegend):

1st = Pythagorean 2nd = Reciprocal 3rd =Algebra 4th = Quotient

OpenStudy (anonymous):

thanks for your help

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