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November 11th, 2018, 12:29 PM   #1
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Question

cos A = 1/3, 0 < A < π/2; sin B = -1/2, 3π/2 < B < 2π

Find tan(A + B)

π = pi

Last edited by skipjack; November 12th, 2018 at 02:19 AM.
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November 11th, 2018, 10:23 PM   #2
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$\begin {align*}
&\tan(a+b) = \\

&\dfrac{\sin(a+b)}{\cos(a+b)} =\\

&\dfrac{\sin(a)\cos(b) + \cos(a)\sin(b)}{\cos(a)\cos(b)-\sin(a)\sin(b)} = \\

&\cos(a) = \dfrac 1 3,~0 < a <\dfrac \pi 2 \Rightarrow \sin(a) = \dfrac{2\sqrt{2}}{3} \\

&\sin(b) = -\dfrac 1 2,~\dfrac{3\pi}{2} < b < 2\pi \Rightarrow \cos(b) = \dfrac{\sqrt{3}}{2} \\

&\tan(a+b) = \dfrac{\dfrac{2\sqrt{2}}{3}\cdot \dfrac{\sqrt{3}}{2} +\dfrac 1 3\cdot \left(-\dfrac 1 2\right)}
{\dfrac 1 3\cdot \dfrac{\sqrt{3}}{2} - \dfrac{2\sqrt{2}}{3} \cdot \left( -\dfrac 1 2\right)} = \\

&\dfrac{2\sqrt{6}-1}{\sqrt{3}+2\sqrt{2}} = \\

&\dfrac{(2\sqrt{6}-1)(\sqrt{3}-2\sqrt{2})}{11} = \\

&\dfrac{8\sqrt{2}-9\sqrt{3}}{11}

\end{align*}$
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November 12th, 2018, 02:04 AM   #3
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tan(A) = 2√2 and tan(B) = -1/√3

tan(A + B) = (2√2 - 1/√3)/(1 + 2√2/√3)
$\hspace{71px}$ = (2√2 - 1/√3)(-1 + 2√2/√3)/(8/3 - 1)
$\hspace{71px}$ = (9√3 - 8√2)/5
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