# $(\sin\theta+\cos\theta)^2=1+\sin2\theta$

49) $(\sin\theta+\cos\theta)^2=1+\sin2\theta$
Left Side:
\begin{align*} (\sin\theta+\cos\theta)^2=\sin^2\theta+2c\cos\theta\sin\theta+cos^2\theta=1+2\cos\theta\sin\theta \end{align*} This can either be $1$ or I can power reduce it. I don't know.

Right Side:
\begin{align*} 1+\sin2\theta=1+2\sin\theta\cos\theta \end{align*}

Thank you!

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$(\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \cos \theta \sin \theta + \cos^2 \theta$. –  Henry T. Horton Jul 15 '12 at 2:13
@HenryT.Horton Can you please explain this rule and/or how you got this answer. Please –  Austin Broussard Jul 15 '12 at 2:15
Please: $$(a+b)^2=a^2+2ab+b^2$$ Multiply out $$(a+b)(a+b)=a(a+b)+b(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2$$ –  Pedro Tamaroff Jul 15 '12 at 2:15
@PeterTamaroff Wow, right. I got that now. –  Austin Broussard Jul 15 '12 at 2:16

Open parentheses and use:

$$(1)\,\,\sin^2x+\cos^2x=1$$

$$(2)\,\,\sin 2x=2\sin x\cos x$$

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Your left hand side isn't good: $(a + b)^2 = a^2 + 2ab + b^2$. After that use $\sin^2(\theta) + \cos^2(\theta) = 1$

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Remember the binomial formula. Importantly, $(a+b)^2\neq a^2+b^2$ ! Rather,

\begin{align*} (\sin\theta+\cos\theta)^2=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta = 1 + 2\sin\theta\cos\theta \end{align*}

Looking at your previous questions, I think you should be a little more careful with the 'simpler' steps in your calculations and double-check those, otherwise you get lost further down.

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