Rewrite with only sin x and cos x sin 2x - cos 2x
@Hero
Just to be clear, you mean \(\sin^2(x) - \cos^2(x)\) or is it \(\sin(2x) - \cos(2x)\)?
Hello again, the one thing i know about this problem is that i should solve it with sum to product formula anddd
Does it say that it has to be solved that way?
\[\sin \left( 2x \right)-\cos \left( 2x \right)\]
it says to rewrite it with only sin x and cos x, so im assuming that i should do it like that
but looking through my sheet, there is nothing on sin a -cos b
I'd say use sum and difference formulas but that might not be the best approach.
the answer choices are: 2sin^2x -2sinx cosx+1 2 sinx 2 sin^2x +sinx cosx -1 2 sin^2x -2sinx cosx -1
so i thought that the sum to product formula was the answer
Hmmm, hang on
You can use sum and difference for this
really?!
Yes because \(\cos(2x) = \cos(x + x)\) \(\sin(2x) = \sin(x + x)\)
having cos(x+x) = cosx cosx + sinx sinx
and sin(x+x) = sinx cosx + cosx sinx
Yes
But remember, you are subtracting
the cos(2x)
so sin(x+x) - cos(x+x)
@Hero
i don't know if i am doing the right thing...
It's the simplest approach
A) 2 sin^2x - 2sinx cosx +1 B) 2 sinx C) 2 sin^2x + 2sinx cosx - 1 D) 2 sin^2x - 2sinx cosx - 1
haha thank you!
are you sure? none of the choices match your final answer
haha
\[\begin{align*} \sin(2x) - \cos(2x) &= \cos(x)\sin(x) + \sin(x) \cos(x) - (\cos(x)\cos(x) + \sin(x)\sin(x) \\&=\sin(2\cos(x) - \cos^2x + \sin^2x \\&= 2\cos(x)\sin(x) - (\cos^2x - \sin^2x) \\&=2\cos(x)\sin(x) - (1 - 2\sin^2x) \\&= 2\sin^2x + 2\cos(x)\sin(x) - 1\end{align*} \]
haha thank you very very much! You really are a Hero :)
I checked it over and you are correct! thank you very much for your time! You are my hero! haha
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