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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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    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. – 2017-02-17
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    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. – 2017-02-17
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    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' – 2017-02-17
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    I have been working on this problem for 2 days. I can't get the right answer in mymathlab. I have tried everything. So, I am not asking for someone to do my homework for me. I have done all the other questions but this one. I can seem to manage it. I factored out a cos^2x and I got (sinxcos^2x+cos^3x). All I can seem to do from here is change the cosine squared x to 1-sinx. But there is where I get stuck. Sorry if I went about asking the question the wrong way. This is the first time I have used a site like this. I am a 40 year old college student trying to get through this section. – 2017-02-17
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    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. – 2017-02-17
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    There are only 8 questions. It is part of a quiz and I have a test on it on Monday. I know it won't downgrade me that much, but I currently have a A in the class and I am afraid of losing too many points on a quiz. I originally got three wrong which brought me to a 68%. That is horrible. I fixed the mistakes on the other two and this is the only one I have left. I am afraid I might get a question like this on the test so I wanted to try to get it down before Monday. I thank you for all your help. – 2017-02-17
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    @HeatherGiovannitti Check https://www.wolframalpha.com/input/?i=sin+xcos%5E4+%2B+sin%5E6+x see if any matches the answer. – 2017-02-17
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    that's not even an equation w h a t a r e y o u d o i n g – 2017-02-17
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    No. It isn't. Yet. I have to solve for the right side – 2017-02-17
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    Does the mymathlab problem tell you to "solve" or to "simplify" the expression $\sin x\cos^4x+\cos^6x$? (And are you sure you've copied the expression *exactly* as it appears there? It would make more sense, to me at least, if you were being asked to simplify $\sin^2x\cos^4x+\cos^6x$.) – 2017-02-17
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    I have to use the trig identities to get it down to it simplest form. We have to find out how to do these first before they put anything on the right side....which would be much easier btw. It is exactly like that on mymathlab. It would make more sense the way you put it, unfortunately it isn't like that. I think I will just ask the professor. Since it is a quiz I probably won't get an answer though. thanks – 2017-02-17
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    If mymathlab is literally saying "solve $\sin x\cos^4x+\cos^6x=$" with nothing on the right hand side, that's a defect in mymathlab, because that's a very weird way of putting it. And if it's really just asking to "simplify" the trig expression, it's not at all clear what it wants as a simplification; either of the already-posted answers is acceptable! – 2017-02-17
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    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. – 2017-02-17
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    Author has posted what they have tried, why downvotes and votes to close? – 2017-02-18
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    @HeatherGiovannitti : do not be disheartened, there are many poeple who will help, you can also help us by providing more info, once I was downvoted 100 points or so, sometimes it just happens, keep doing math. – 2017-02-18

3 Answers 3

4

It is already in simplest terms

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    I don't normally upvote answers like this, but barring any other context from the OP, I agree with this answer. – 2017-02-17
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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}