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Find the exact value of each expression:

1) $\sin{(-\frac{\pi}{2} +\frac{\pi}{3})}$ -For this question, it would appear as though you could use the addition compound angle formula $\sin{(A+B)}=\sin{A}\cos{B}+\sin{B}\cos{A}$, however due to the $-$ sign in front of the $\frac{\pi}{2}$, I am not sure if this is still considered to be apart of the special triangles. I know that $\frac{\pi}{2}$ (90 degrees) is. By the way, these questions are to be in radian measure.

2) $\tan{ (\frac{7\pi}{12})}$ -I think this one can be split into $\tan{(\frac{3\pi}{12} + \frac{4\pi}{12})}$ and get $\tan{(\frac{\pi}{4} + \frac{\pi}{3})}$ and then input the values into $\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$

3) $2\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}$ -This one I am not sure about where to begin. I am not sure which identity I would use here since this could be $\sin{a}\cos{b}$ could be used with either addition or subtraction.

If someone could help me out with these questions, that would be great!

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  • $\begingroup$ For 1), you can also use $\sin(A - B) = \sin A \cos B - \cos A \sin B$ with $A = \frac{\pi}{3}$ and $B = \frac{\pi}{2}$. $\endgroup$ – N. F. Taussig Apr 3 '15 at 18:33
  • $\begingroup$ Oh that's a great idea! Thanks for the tip. $\endgroup$ – Ash Apr 3 '15 at 18:35
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    $\begingroup$ So sin(pi/3-pi/2)=sinpi/3cospi/2-sinpi/2cospi/3? After this I need to use the special triangles. sinpi/3 gives (root3/2) but what of cospi/2 and sinpi/2? $\endgroup$ – Ash Apr 3 '15 at 18:53
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    $\begingroup$ Okay, well I'm not sure if I did this right but it came to (root3/2)(0)-(1)(1/2) $\endgroup$ – Ash Apr 3 '15 at 19:27
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    $\begingroup$ Equaling -1/2 since anything multiplied by 0 equals zero. $\endgroup$ – Ash Apr 3 '15 at 19:28
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The first question can be handled with the difference of angles formula $$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$$ where $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{2}$ since $-\frac{\pi}{2} + \frac{\pi}{3} = \frac{\pi}{3} - \frac{\pi}{2}$. Substitution yields \begin{align*} \sin\left(-\frac{\pi}{2} + \frac{\pi}{3}\right) & = \sin\left(\frac{\pi}{3} - \frac{\pi}{2}\right)\\ & = \sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{3}\right)\\ & = \frac{\sqrt{3}}{2} \cdot 0 - 1 \cdot \frac{1}{2}\\ & = 0 - \frac{1}{2}\\ & = -\frac{1}{2} \end{align*} Check:
\begin{align*} \sin\left(-\frac{\pi}{2} + \frac{\pi}{3}\right) & = \sin\left(-\frac{\pi}{6}\right)\\ & = -\sin\left(\frac{\pi}{6}\right)\\ & = -\frac{1}{2} \end{align*} since the sine function is negative in the fourth quadrant.

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HINT:

Recount $\sin(A+B)$ formula holds true for all finite values of $A,B$

$(1)$ How about $A=-\dfrac\pi2, B=\dfrac\pi3$

$(2)$ Proceed with the formula

$(3)$ Put $A=B$ in $\sin(A+B)$ formula

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For 3rd one, we can use this formula $$\sin(2A)=2\sin(A)\cos(A).$$ So the answer will be $1/\sqrt{2}$.

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