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Mathematics 7 Online
OpenStudy (anonymous):

Rewrite with only sin x and cos x sin 2x - cos 2x

OpenStudy (anonymous):

@Hero

hero (hero):

Just to be clear, you mean \(\sin^2(x) - \cos^2(x)\) or is it \(\sin(2x) - \cos(2x)\)?

OpenStudy (anonymous):

Hello again, the one thing i know about this problem is that i should solve it with sum to product formula anddd

hero (hero):

Does it say that it has to be solved that way?

OpenStudy (anonymous):

\[\sin \left( 2x \right)-\cos \left( 2x \right)\]

OpenStudy (anonymous):

it says to rewrite it with only sin x and cos x, so im assuming that i should do it like that

OpenStudy (anonymous):

but looking through my sheet, there is nothing on sin a -cos b

hero (hero):

I'd say use sum and difference formulas but that might not be the best approach.

OpenStudy (anonymous):

the answer choices are: 2sin^2x -2sinx cosx+1 2 sinx 2 sin^2x +sinx cosx -1 2 sin^2x -2sinx cosx -1

OpenStudy (anonymous):

so i thought that the sum to product formula was the answer

hero (hero):

Hmmm, hang on

hero (hero):

You can use sum and difference for this

OpenStudy (anonymous):

really?!

hero (hero):

Yes because \(\cos(2x) = \cos(x + x)\) \(\sin(2x) = \sin(x + x)\)

OpenStudy (anonymous):

having cos(x+x) = cosx cosx + sinx sinx

OpenStudy (anonymous):

and sin(x+x) = sinx cosx + cosx sinx

hero (hero):

Yes

hero (hero):

But remember, you are subtracting

hero (hero):

the cos(2x)

OpenStudy (anonymous):

so sin(x+x) - cos(x+x)

OpenStudy (anonymous):

@Hero

OpenStudy (anonymous):

i don't know if i am doing the right thing...

hero (hero):

It's the simplest approach

OpenStudy (anonymous):

A) 2 sin^2x - 2sinx cosx +1 B) 2 sinx C) 2 sin^2x + 2sinx cosx - 1 D) 2 sin^2x - 2sinx cosx - 1

OpenStudy (anonymous):

haha thank you!

OpenStudy (anonymous):

are you sure? none of the choices match your final answer

OpenStudy (anonymous):

haha

hero (hero):

\[\begin{align*} \sin(2x) - \cos(2x) &= \cos(x)\sin(x) + \sin(x) \cos(x) - (\cos(x)\cos(x) + \sin(x)\sin(x) \\&=\sin(2\cos(x) - \cos^2x + \sin^2x \\&= 2\cos(x)\sin(x) - (\cos^2x - \sin^2x) \\&=2\cos(x)\sin(x) - (1 - 2\sin^2x) \\&= 2\sin^2x + 2\cos(x)\sin(x) - 1\end{align*} \]

OpenStudy (anonymous):

haha thank you very very much! You really are a Hero :)

OpenStudy (anonymous):

I checked it over and you are correct! thank you very much for your time! You are my hero! haha

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