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Verify that (sin x)(tan x cos x – cot x cos x) = 1 – 2 cos2x

Trigonometry
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did you write this down correctly
yes
its very confusing

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Other answers:

are you sure the last one shouldn't be \[1 - 2 \cos^{2}(x) \] instead of 1 - 2 cos 2x
yes that is what i meant sorry it didnt square it
Ok try starting by rewriting tan x as sin x / cos x and cot x as cos x / sin x
\[(\sin x)(\frac{ \sin x }{ \cos x } * \cos x - \frac{ \cos x }{ \sin x } * \cos x) = 1 - 2 \cos^{2} x\]
ok so cos^2x/sinx ?
at the end ?
it would become \[(\sin x)(\sin x - \frac{ \cos^{2} x }{ \sin x }) = 1 - 2 \cos^{2} x\] then \[\sin^{2} x - \cos^{2} x = 1 - 2 \cos^{2} x\]
last step would just be add 2 cos^2(x) to both sides
i dont understand where the 2cos^2x came from
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
and we have the identity \[\sin^{2} x + \cos^{2} x = 1\]
oh ok i understand! thanks! do you think you can help me out with another one?
sure but they recommend you close this question and start another one
brb though
ok so ill open another

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