# Trig identities

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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Is that $(1+\sin t)^{2} + \cos^{2} t$ or $(1+\sin t)^{2} + \cos^{2t}$? Same question for the others, as well. –  Jack Henahan Jun 15 '11 at 18:33
While my edit might not reflect the intended expression, it should give Adam a starting place from which to edit. –  The Chaz 2.0 Jun 15 '11 at 18:37
I got it, I was 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) –  Adam Jun 15 '11 at 18:43

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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I don't get how sin^2 turns into 2sin or whatever happened. –  Adam Jun 15 '11 at 18:40
@Adam: It doesn't turn into 2sin but $2(1+\sin{t})$ –  user9413 Jun 15 '11 at 18:49

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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Chris, keep checking back, as we aren't totally sure as to what the intended expression is/was! –  The Chaz 2.0 Jun 15 '11 at 18:39
@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. –  Arturo Magidin Jun 15 '11 at 18:45