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anonymous
 4 years ago
Verify that (sin x)(tan x cos x – cot x cos x) = 1 – 2 cos2x
anonymous
 4 years ago
Verify that (sin x)(tan x cos x – cot x cos x) = 1 – 2 cos2x

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you write this down correctly

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes that is what i meant sorry it didnt square it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok try starting by rewriting tan x as sin x / cos x and cot x as cos x / sin x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\sin x)(\frac{ \sin x }{ \cos x } * \cos x  \frac{ \cos x }{ \sin x } * \cos x) = 1  2 \cos^{2} x\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0last step would just be add 2 cos^2(x) to both sides

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont understand where the 2cos^2x came from

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there 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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure but they recommend you close this question and start another one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so ill open another
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