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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!

  • 2
    $(\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \cos \theta \sin \theta + \cos^2 \theta$.2012-07-15
  • 0
    @HenryT.Horton Can you please explain this rule and/or how you got this answer. Please2012-07-15
  • 1
    Please: $$(a+b)^2=a^2+2ab+b^2$$ Multiply out $$(a+b)(a+b)=a(a+b)+b(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2$$2012-07-15
  • 0
    @PeterTamaroff Wow, right. I got that now.2012-07-15

3 Answers 3

3

Open parentheses and use:

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

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

3

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.

2

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