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 Trigonometry Trigonometry Math Forum

 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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post harryp Algebra 2 September 8th, 2015 08:11 PM mared Trigonometry 3 July 2nd, 2014 08:09 AM kingkos Algebra 5 November 24th, 2012 11:40 PM kevinman Algebra 8 March 8th, 2012 05:53 AM

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