Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $2\sin$

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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: $$\begin{align*} (1+\sin t)^2 + \cos^2 t &= \Bigl( 1 + 2\sin t + \sin^2 t\Bigr) + \cos^2 t\\ &= 1 + 2\sin t + \Bigl( \sin^2 t + \cos^2 t\Bigr)\\ &= 1 + 2\sin t + 1\\ & = 2 + 2\sin t\\ & = 2 (1 + \sin t). \end{align*}$$

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.