 My Math Forum How do I prove this seemingly simple trigonometric identity
 User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 December 11th, 2014, 07:15 PM #1 Senior Member   Joined: Aug 2014 From: India Posts: 343 Thanks: 1 How do I prove this seemingly simple trigonometric identity $\displaystyle a=\sinθ+\sinϕ$ $\displaystyle b=\tanθ+\tanϕ$ $\displaystyle c=\secθ+\secϕ$ Show that, $\displaystyle 8bc=a[4b^2 + (b^2-c^2)^2]$ I tried to solve this for hours and have gotten nowhere. Here's what I've got so far: $\displaystyle \\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})$ $\displaystyle \\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+ \cos(\theta-\phi)}$ $\displaystyle \\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi )+\cos(\theta-\phi)}$ $\displaystyle \\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]} {\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca +b)}$ Last edited by skipjack; December 12th, 2014 at 12:44 AM. December 12th, 2014, 02:59 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 \displaystyle \begin{align*}4\cos^2(\theta)\cos^2(\phi)b^2 &= 4\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))^2 \\ &= 4(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^2 \\ &= 4\sin^2(\theta + \phi) \\ &= 16\sin^2((\theta + \phi)/2)\cos^2((\theta + \phi)/2) \\ &= 16(1 - \cos^2((\theta + \phi)/2))\cos^2((\theta + \phi)/2) \\ &= 16\cos^2((\theta + \phi)/2) - 16\cos^4((\theta + \phi)/2)\end{align*} \displaystyle \begin{align*}c^2 - b^2 &= \sec^2\theta + \sec^2\phi + 2\sec(\theta)\sec(\phi) - (\tan^2\theta + \tan^2\phi + 2\tan(\theta)\tan(\phi)) \\ &= 2 + 2\sec(\theta)\sec(\phi) - 2\tan(\theta)\tan(\phi)\end{align*} \displaystyle \begin{align*}\cos(\theta)\cos(\phi)(c^2 - b^2) &= 2(\cos(\theta)\cos(\phi) + 1 - \sin(\theta)\sin(\phi)) \\ &= 2(1 + \cos(\theta + \phi)) \\ &= 4\cos^2((\theta + \phi)/2)\end{align*} $\displaystyle \cos^2(\theta)\cos^2(\phi)(b^2 - c^2)^2 = 16\cos^4((\theta + \phi)/2)$ $\displaystyle \cos^2(\theta)\cos^2(\phi)(4b^2 + (b^2 - c^2)^2) = 16\cos^2((\theta + \phi)/2)$ \displaystyle \begin{align*}8\cos^2(\theta)\cos^2(\phi)bc/a &= 8\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))(\sec(\theta) + \sec(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= 8(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= 8\sin(\theta + \phi)(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= \frac{8(2\sin((\theta + \phi)/2)\cos((\theta + \phi)/2))(2\cos((\theta + \phi)/2)\cos((\theta - \phi)/2))}{2\sin((\theta + \phi)/2)\cos((\theta - \phi)/2)} \\ &= 16\cos^2((\theta+\phi)/2)\end{align*} The desired result follows. Tags identity, prove, seemingly, simple, trigonometric 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 matemagia Trigonometry 4 August 30th, 2012 06:40 PM OriaG Trigonometry 2 August 13th, 2012 03:35 PM afrim Trigonometry 8 October 1st, 2011 01:59 AM coolguyadarsh Trigonometry 1 April 23rd, 2011 08:32 AM chnixi Algebra 1 August 19th, 2009 03:52 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      