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binarymimicBest ResponseYou've already chosen the best response.0
did you write this down correctly
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
are you sure the last one shouldn't be \[1  2 \cos^{2}(x) \] instead of 1  2 cos 2x
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
yes that is what i meant sorry it didnt square it
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
Ok try starting by rewriting tan x as sin x / cos x and cot x as cos x / sin x
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
\[(\sin x)(\frac{ \sin x }{ \cos x } * \cos x  \frac{ \cos x }{ \sin x } * \cos x) = 1  2 \cos^{2} x\]
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
ok so cos^2x/sinx ?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
it would become \[(\sin x)(\sin x  \frac{ \cos^{2} x }{ \sin x }) = 1  2 \cos^{2} x\] then \[\sin^{2} x  \cos^{2} x = 1  2 \cos^{2} x\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
last step would just be add 2 cos^2(x) to both sides
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
i dont understand where the 2cos^2x came from
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
there is already a 2 cos^2 x in the problem. the last step is to add 2 cos^2 x to both sides this way the right side becomes: 1 and the left side becomes: sin^2 x + cos^2 x
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
and we have the identity \[\sin^{2} x + \cos^{2} x = 1\]
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
oh ok i understand! thanks! do you think you can help me out with another one?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
sure but they recommend you close this question and start another one
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
ok so ill open another
 one year ago
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