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Trigonometry 8 Online
OpenStudy (anonymous):

cos^2(2x)+cos^2(x)=1 -π<=x<=π/3

OpenStudy (xapproachesinfinity):

what did you do so far?

OpenStudy (xapproachesinfinity):

Hint \(cos(2x)=cos^2x-sin^2x\)

OpenStudy (anonymous):

but what i need to do with the ^2?

OpenStudy (phi):

better, cos(2x)= 2 cos^2(x) - 1 and square that

OpenStudy (xapproachesinfinity):

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

OpenStudy (xapproachesinfinity):

that distribution you did was incorrect! you know? it is not con^4x-sin^4x

OpenStudy (xapproachesinfinity):

\((a-b)^2=a^2-\color{magenta}{2ab}+b^2\) you forgot the term in the middle

OpenStudy (anonymous):

Yea i deleted this.. but its not cos2x, its cos^2(2x).. so wahat i need to do now?

OpenStudy (xapproachesinfinity):

But anyways use the \(2cos^2x-1\)

OpenStudy (anonymous):

what*

OpenStudy (xapproachesinfinity):

square that \(cos^2(2x)=(2cos^2x-1)^2\)

OpenStudy (xapproachesinfinity):

you got it ?

OpenStudy (anonymous):

oaa ok... thx :) let me try

OpenStudy (xapproachesinfinity):

okay

OpenStudy (anonymous):

|dw:1408812735899:dw| so now i got this.. how i continue?

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