# Trigonometric factoring

Very next question, no idea what to do...

I am suppose to factor $2\sin^2x + 3\sin x+1$ .

I figure this is pretty simple so I do $(2\sin x)(2\sin x)+3 \sin x+1$ .

For some reason this is incorrect (not sure why) and they give $(2\sin x+1)(\sin x+1)$ which I did verify gives me the original but I don't know how to get there.

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What you did is not incorrect (apart from a 2), but it is not a factorization of the whole expression: you are supposed to get a product of two factors $2\sin^2x + 3\sin x+1 = (something)(anotherthing)$ and your approach didn't lead you there. – leonbloy Jun 15 '11 at 21:38
@Adam: At some time, you learned how to "factor" $2y^2+3y+1$. You got $(2y+1)(y+1)$. Now let $y=\sin x$ and remember that $\sin^2 x$ means $(\sin x)^2$. You end up with $(2\sin x +1)(\sin x +1)$. Maybe check by multiplying out. – André Nicolas Jun 15 '11 at 21:40
@Adam: You are having trouble with basic mathematical conventions. For example, $2\sin^2 x$ means $2\times (\sin x)^2$. From what you wrote, you interpreted it as $(2 \times \sin x)^2$. Now I will admit that the notation $\sin^2 x$ for the square of the sine of $x$, is not best possible. But it is nearly universally used, so one has to become accustomed to it. – André Nicolas Jun 15 '11 at 22:38
Yeah I know I am horrible at math, I should probably change my major out of engineering. – Adam Jun 15 '11 at 22:41
Unfortunately (or fortunately if you end up liking math), you're going to be doing a LOT and I mean A LOT of math. Half your classes if not more are going to be math classes if you're degree is in engineering lol. – OghmaOsiris Jun 15 '11 at 23:39

Think of $2sin^2(x) + 3sin(x) + 1$ as a whole function similar to $2x^2 + 3x + 1$ and see if it makes sense now.

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I tried that and I have no idea what to do, seems like I forgot everything I learned about math a couple months ago. – Adam Jun 15 '11 at 21:33
Well, can you factor $2x^2 + 3x + 1$? – OghmaOsiris Jun 15 '11 at 21:34
I am having trouble thinking about how to do it since I can't complete the square or any of the obvious methods I know. Trial and error I suppose. I got (2x+1)(x+1) with trial and error. – Adam Jun 15 '11 at 21:37
Well, what are the factors of 1? There are none, so obviously the factored form will be $($_$+ 1)($_$+ 1)$ since both the signs in the polynomial are positive. So what combinations will give you a $3x$ when you add them together and also give you $2x^2$? $2x$ and $1x$ (or just $x$). So you end up with what? – OghmaOsiris Jun 15 '11 at 21:40
Thanks I got it, unfortunately I got stuck on the very next problem. – Adam Jun 15 '11 at 21:44

Looking at: $2\sin^2x + 3\sin x+1$

Note that $\sin^2x$ means nothing more than $(\sin x)^2$, so you can rewrite your expression as $2(\sin x)^2 + 3(\sin x) + 1$. Then, as others have suggested, set $y = \sin x$ --> then substituting into the original expression, we have $2y^2+ 3y + 1$. Once you factor that, remember to replace every $y$ with $\sin x$, and you should have the factored expression that matches your book's answer.

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Yes, of course, @Babak ! (Sorry I didn't catch your message earlier!) – amWhy Jul 10 '13 at 0:59
Which expression do you use when you are scaring cause of very horrible things and you feel this state on your skin? I want this word. I saw one of Judy Foster's recent work and somewhere I felt she told a word which I didn't catch. – Babak S. Jul 10 '13 at 1:25
Let me think about it...I know what you mean, I simply can't recall this moment what term Jodie Foster used. Sometimes, people say: "I was so scared, that my skin was crawling!" – amWhy Jul 10 '13 at 1:30
Bingo! I think she said something like it. Thanks so much. :-) – Babak S. Jul 10 '13 at 1:32

Have you tried multiplying out the book answer to see what you get? Again, the fact that you have $\sin x$ is not important: $2y^2+3y+1=(2y+1)(y+1)$. Your answer is not in a factored form, nor is it correct as $(2\sin x)(2\sin x)=4\sin^2 x$

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