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How does this: $$(3+\cos x)(4x\cos x+4\sin x)-(4x\sin x)(-\sin x)$$

simplify to this: $$4\sin x\cos x +12x\cos x+12\sin x+4x$$ ?

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  • $\begingroup$ product rule and distribution $\endgroup$
    – izzzi
    Oct 18, 2016 at 1:19

2 Answers 2

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Multiply and combine terms.

Use $\sin^2(x)+\cos^2(x) = 1$.

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$$(3+\cos x)(4x \cos x+4 \sin x)−(4x \sin x)(−\sin x)$$ $$= 12x \cos x + 12 \sin x + 4x \cos^2x + 4 \sin x \cos x + 4x \sin^2x$$ $$= 12x \cos x + 12 \sin x + 4 \sin x \cos x + (4x \sin^2x + 4x \cos^2x) $$ $$= 4 \sin x \cos x + 12x \cos x + 12 \sin x + 4x $$

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    $\begingroup$ @suomynonA, Thanks for the edit. I was in the process of re-editing it to use standard math fonts. $\endgroup$
    – blackpen
    Oct 18, 2016 at 1:36
  • $\begingroup$ @DustinMorrow: Edits should not make substantive changes to posts. Work on increasing your reputation so that you will be able to Comment on the posts of others and point out what you believe to be substantive errors, or else post your own Answer if you believe you have one that is substantively different. $\endgroup$
    – hardmath
    Oct 18, 2016 at 1:44

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