How do you verify $sin x + cos x = (2sin^2 x - 1) /(sin x -cos x)$ ?

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Answer 1 · 2016-07-09T20:58:31

$sinx + cosx =(2(1 - cos^2x) - 1)/(sinx - cosx)$

$sinx + cosx= (2 - 2cos^2x - 1)/(sinx - cosx$

$sinx + cosx = (1 - 2cos^2x)/(sinx - cosx)$

Here it looks like we're at a dead end, but in fact we're not.

$(sinx + cosx)(sinx - cosx) = 1 - 2cos^2x$

$sin^2x - cos^2x = 1 - 2cos^2x$

$1 - cos^2x - cos^2x = 1 - 2cos^2x$

$1 - 2cos^2x = 1 - 2cos^2x -> "Identity proved!"$

Hopefully this helps!

Answer 2 · 2016-07-10T02:02:17

$RHS=(2sin^2x-1)/(sinx-cosx)$

$=(2sin^2x-(cos^2x+sin^2x))/(sinx-cosx)$

$=(2sin^2x-cos^2x -sin^2x)/(sinx-cosx)$

$=(sin^2x-cos^2x)/(sinx-cosx)$

$=((cancel(sinx-cosx))(sinx+cosx))/(cancel(sinx-cosx))$

$=(sinx+cosx)=LHS$

How do you verify sin x + cos x = (2sin^2 x - 1) /(sin x -cos x)sin x + cos x = (2sin^2 x - 1) /(sin x -cos x)?