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 $2\sin$
 A: 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:
$$\begin{align*}
(1+\sin t)^2 + \cos^2 t &= \Bigl( 1 + 2\sin t + \sin^2 t\Bigr) + \cos^2 t\\
&= 1 + 2\sin t + \Bigl( \sin^2 t + \cos^2 t\Bigr)\\
&= 1 + 2\sin t + 1\\
& = 2 + 2\sin t\\
& = 2 (1 + \sin t).
\end{align*}$$
A: 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.
A: You can write this as $$ (1+\sin{t})^{2} + \cos^{2}{t} = 1 + 2\sin{t} + \sin^{2} + \cos^{2}{t} = 2 + 2 \sin{t}=2\cdot (1+\sin{t})$$
