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Use Trig Identities to solve $\sin x \cos^4 x+\cos^6 x$=

My Steps: I am trying to solve this problem with Trig Identities. I factored out a $\cos^2x$ and then changed the $\cos^2x$ to $1-\sin^2x$. I am having most of my problems with the simple algebra that I can't remember from high school. So far I brought it down to $(1-\sin^2x)(\sin x-\sin^2x+\cos x)$. Did I do that part right? Also, where would I go from here?

I found the answer!

All I had to do was factor out a cos^4x

Thank you all!

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closed as off-topic by Alfred Yerger, Leucippus, Juniven, user91500, TheGeekGreek Feb 18 '17 at 7:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alfred Yerger, Leucippus, Juniven, user91500, TheGeekGreek
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You're probably getting downvoted for a few reasons. 1) There's no context or motivation for this, and no effort on your part, 2) it already looks like it is in simplest terms. Most people on this site would probably not do anything further to this without a specific reason to want to. $\endgroup$ – Alfred Yerger Feb 17 '17 at 18:36
  • $\begingroup$ I guess I thought this was a site to get math help. I have no one else to ask. As far as it being in its simplest form, I really don't know. We are doing trig Identities and my job is to break these down into simpler forms using those identities. If people don't want to help there is nothing I can do about that. Oh well. $\endgroup$ – Heather Giovannitti Feb 17 '17 at 18:39
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    $\begingroup$ Right, it is a help site, but it's not a 'do-your-homework-for-you' site. Tell us where you're stuck, what identities you have to work with, etc. What is the end goal of this simplification? I'd damn well say this IS simplest. The best I can think of is to try using formulas to reduce powers on cosines, but all that's going to do is make more terms. Not really 'simpler' $\endgroup$ – Alfred Yerger Feb 17 '17 at 18:41
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    $\begingroup$ It's okay. Nobody begrudges you not knowing the usual way things work around here. Edit it in your thoughts from that last comment, and people will probably remove their downvotes. For what it's worth, if nobody knows what the software wants, I wouldn't stress too hard about one problem. If you can do all the others, then you probably understand the main gist of things. One trick question is probably okay, as long as your grade isn't jeopardized by it. $\endgroup$ – Alfred Yerger Feb 17 '17 at 18:56
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    $\begingroup$ Ask your instructor what is expected here. Mymathlab can be very specific in some cases as to what it expects. If you can use some parens and explicitly place a multiplication sign when entering to see what the expected answer is. $\endgroup$ – scrappedcola Feb 17 '17 at 21:29
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It is already in simplest terms

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  • $\begingroup$ I don't normally upvote answers like this, but barring any other context from the OP, I agree with this answer. $\endgroup$ – Alfred Yerger Feb 17 '17 at 18:43
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i have another idea: writing $$\cos(x)^4(\sin(x)+\cos(x)^2)=\cos(x)^4(-\sin(x)^2+\sin(x)+1)$$

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Maybe, the OP wants to get rid of powers:

$$ \left\{\begin{array}{rcl} \ds{\cos^{4}\pars{x}} & \ds{=} & \ds{\cos\pars{4x} + 4\cos\pars{2x} + 3 \over 8} \\[2mm] \ds{\cos^{6}\pars{x}} & \ds{=} & \ds{\cos\pars{6x} + 6\cos\pars{4x} + 15\cos\pars{2x} + 10 \over 32} \end{array}\right. $$


\begin{align} &\sin\pars{x}\cos^{4}\pars{x} + \cos^{6}\pars{x} \\[5mm] = &\ \frac{3 \sin (x)}{8}+\frac{15}{32} \cos (2 x)+\frac{3}{16} \cos (4 x)+\frac{1}{32} \cos (6 x)+\frac{1}{2} \sin (x) \cos (2 x)+\frac{1}{8} \sin (x) \cos (4 x)+\frac{5}{16} \end{align}

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