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I need to perform the indicated operation and simplify $(1+\sin t)^{2} + \cos^{2} t$

The book is telling me that it turns into $1 + 2\sin^2t + \cos^2t$, how is is possible? Basic math tells me that 2(3) is equal to six and that $3^2 = 9$ so there is no way that $\sin^2$ can be turned into $2\sin$

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    I got it, I w$a$s just doing the math completely wrong, I forgot how to square something in parentheses. I constantly make these kinds of mistakes no matter how hard I try not, especially on tests. I doubt I will make it far in calculus (which I am sure most people here consider easy high school math)2011-06-15

3 Answers 3

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You can write this as $ (1+\sin{t})^{2} + \cos^{2}{t} = 1 + 2\sin{t} + \sin^{2} + \cos^{2}{t} = 2 + 2 \sin{t}=2\cdot (1+\sin{t})$

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    @Adam: It doesn't turn into 2sin but $2(1+\sin{t})$2011-06-15
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I don't understand what your book is suggesting.

If you first expand the squared binomial, remembering that $(a+b)^2 = a^2 + 2ab + b^2$, we have $(1+\sin t)^2 = 1^2 + 2\times 1 \times \sin t + \sin^2 t = 1 + 2\sin t + \sin^2 t.$ Then, use the fact that $\sin^2 t + \cos^2 t = 1$. So we have: $\begin{align*} (1+\sin t)^2 + \cos^2 t &= \Bigl( 1 + 2\sin t + \sin^2 t\Bigr) + \cos^2 t\\ &= 1 + 2\sin t + \Bigl( \sin^2 t + \cos^2 t\Bigr)\\ &= 1 + 2\sin t + 1\\ & = 2 + 2\sin t\\ & = 2 (1 + \sin t). \end{align*}$

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If the question is exactly as you've written it, your book is wrong. By multiplying out the bracket:

$(1+\sin t)^2 + \cos^2t = 1 + 2\sin t + \sin^2t + \cos^2 t$

and this further simplifies to

$2(1+\sin t)$

However, you seem to think (and correct me if I'm wrong) that $\sin$ and $\sin^2$ have a meaning independent of their argument, $t$, so it's also possible that you're confused by what the question is asking you to do.

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    @The Chaz: The OP himself rewrote it as `(1 + sin t)^2 + cos^{2}t`, so I think it's pretty clear that is what he meant.2011-06-15