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Precalculus 13 Online
OpenStudy (anonymous):

Rewrite with only sin x and cos x. sin 2x - cos x Options are: 2 sin x cos^2x sin x cos x (2 sin x - 1) 2 sin x

OpenStudy (anonymous):

Tell me what you get when you expand \[\sin(2x)\] Use your double angle results. \[\sin(2x)=\sin(x+x)\] \[\sin(x+x)=?\]

OpenStudy (anonymous):

Doesn't sin 2x = 2 sin x cos x?

OpenStudy (anonymous):

Does that have anything to do with this problem?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

That's exactly right.

OpenStudy (anonymous):

You substitute 2sinxcosx with sin(2x).

OpenStudy (anonymous):

so factorise 2sinxcosx-cosx for me please.

OpenStudy (anonymous):

Well since there is cos x on both side of the equation would you cancel them and then you are left with just 2 sin x?

OpenStudy (anonymous):

Do you know what factorising is?

OpenStudy (anonymous):

For example, can you factorise: \[x^2+x\]?

OpenStudy (anonymous):

would it be x(x+1)

OpenStudy (anonymous):

Correct! That's how you would go about factorising this: 2sinxcosx-cosx

OpenStudy (anonymous):

Take the common factor in both terms and then you can factorise it.

OpenStudy (anonymous):

Would the common factor would be cos x?

OpenStudy (anonymous):

Yes!

OpenStudy (anonymous):

So the answer would be cos x(2 sin x-1)?

OpenStudy (anonymous):

Correct. Well Done.

OpenStudy (anonymous):

Thank you for your help! :)

OpenStudy (anonymous):

No worries.

OpenStudy (anonymous):

useful to know sin or cos shifted a certain amount will equal cos or sin. so in other words an example sin(x+a)=cos (x)

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