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 November 11th, 2018, 12:29 PM #1 Newbie   Joined: Nov 2018 From: United States Posts: 2 Thanks: 0 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.
 November 11th, 2018, 10:23 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 \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*}
 November 12th, 2018, 02:04 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2202 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

 Tags cosa, find, sinb, tan(a+b), tana

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