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 December 18th, 2017, 09:50 AM #1 Member   Joined: Oct 2012 Posts: 59 Thanks: 0 prove Prove that : sin^2(x)*(tan^2(x)+3)/(1-tan^2(x))=sec(2x)-cos^2(x)
 December 18th, 2017, 11:12 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,692 Thanks: 976 Math Focus: Elementary mathematics and beyond It's not an identity.
 December 18th, 2017, 04:57 PM #3 Global Moderator   Joined: Dec 2006 Posts: 18,416 Thanks: 1462 LHS = sin²(x)(sin²(x)/cos²(x) + 3)/(1 - sin²(x)/cos²(x)) $\ \ \ \ \ \$ = sin²(x)(sin²(x) + 3cos²(x))/(cos²(x) - sin²(x)) $\ \ \ \ \ \$ = (1 - cos²(x))(1 + 2cos²(x))/cos(2x) $\ \ \ \ \ \$ = (1 - cos²(x)(1 + 2cos²(x) - 2))/cos(2x) $\ \ \ \ \ \$ = (1 - cos²(x)(cos(2x)))/cos(2x) $\ \ \ \ \ \$ = sec(2x) - cos²(x) Thanks from fahad nasir
 December 29th, 2017, 06:31 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,692 Thanks: 976 Math Focus: Elementary mathematics and beyond \begin{align*}\sin^2(x)\frac{\tan^2(x)+3}{1-\tan^2(x)}&=-\sin^2(x)\frac{\sec^2(x)+2}{\sec^2(x)-2} \\ &=(\cos^2(x)-1)\frac{\sec^2(x)+2}{\sec^2(x)-2} \\ &=\frac{1+2\cos^2(x)-\sec^2(x)-2}{\sec^2(x)-2} \\ &=\frac{2\cos^2(x)-1-\sec^2(x)}{\sec^2(x)-2} \\ &=\frac{\cos(2x)-\sec^2(x)}{\sec^2(x)-2} \\ &=\frac{\cos^2(x)\cos(2x)-1}{1-2\cos^2(x)} \\ &=-\cos^2(x)+\sec(2x)\end{align*} Thanks from fahad nasir
 January 6th, 2018, 04:08 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,692 Thanks: 976 Math Focus: Elementary mathematics and beyond \begin{align*}\sin^2(x)\frac{\tan^2(x)+3}{1-\tan^2(x)}&=\sin^2(x)\frac{2\tan(x)}{1-\tan^2(x)}\frac{\tan^2(x)+3}{2\tan(x)} \\ &=\sin^2(x)\tan(2x)(\tan(x)+3\cot(x))/2 \\ &=\sin^2(x)\sin(2x)\sec(2x)(\tan(x)+3\cot(x))/2 \\ &=\sin^2(x)\sin(x)\cos(x)\sec(2x)(\tan(x)+3\cot(x) ) \\ &=\sin^2(x)\sec(2x)(\sin^2(x)+3\cos^2(x)) \\ &=\sin^2(x)\sec(2x)(1+2\cos^2(x)) \\ &=(1-\cos^2(x))\sec(2x)(1+2\cos^2(x)) \\ &=(1-\cos^2(x))(\sec(2x)+2\cos^2(x)\sec(2x)) \\ &=\sec(2x)+2\cos^2(x)\sec(2x)-\cos^2(x)\sec(2x)-2\cos^4(x)\sec(2x) \\ &=\sec(2x)+\cos^2(x)\sec(2x)-2\cos^4(x)\sec(2x) \\ &=\sec(2x)+\cos^2(x)\sec(2x)(1-2\cos^2(x)) \\ &=\sec(2x)-\cos^2(x)\sec(2x)(2\cos^2(x)-1) \\ &=\sec(2x)-\cos^2(x)\sec(2x)\cos(2x) \\ &=\sec(2x)-\cos^2(x)\end{align*}
 January 7th, 2018, 03:35 PM #6 Senior Member   Joined: Oct 2009 Posts: 180 Thanks: 70 You can very easily and mechanically prove a LOT of trig statements using the tangent half-angle formulas. https://en.wikipedia.org/wiki/Tangen...-angle_formula What it accomplishes is it expresses any value with different functions cos(x), sin(x), sec(x), tan(x), etc as a value using only a parameter t. As such you have reduced the problem to simply showing an equality of polynomials, which is trivial. It's definitely not the shortest way to solve this problem, but it's a problem that requires no thought and it always works.
 January 7th, 2018, 10:13 PM #7 Global Moderator   Joined: Dec 2006 Posts: 18,416 Thanks: 1462 It doesn't "always work". For example, it wouldn't show that sin($\pi$/6 + x) + sin($\pi$/6 - x) ≡ cos(x).
January 8th, 2018, 02:54 AM   #8
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Quote:
 Originally Posted by skipjack It doesn't "always work". For example, it wouldn't show that sin($\pi$/6 + x) + sin($\pi$/6 - x) ≡ cos(x).
Come on, read the post. I said it helped in a LOT of trig problems, not all of them. And I said it always worked in specific situations, where you have the functions sin(x), cos(x) etc.

 January 8th, 2018, 06:35 AM #9 Global Moderator   Joined: Dec 2006 Posts: 18,416 Thanks: 1462 Of course, but "etc." is a way of avoiding being specific.

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