December 18th, 2017, 08:50 AM  #1 
Member Joined: Oct 2012 Posts: 61 Thanks: 0  prove
Prove that : sin^2(x)*(tan^2(x)+3)/(1tan^2(x))=sec(2x)cos^2(x)

December 18th, 2017, 10:12 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
It's not an identity.

December 18th, 2017, 03:57 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
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) 
December 29th, 2017, 05:31 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
$$\begin{align*}\sin^2(x)\frac{\tan^2(x)+3}{1\tan^2(x)}&=\sin^2(x)\frac{\sec^2(x)+2}{\sec^2(x)2} \\ &=(\cos^2(x)1)\frac{\sec^2(x)+2}{\sec^2(x)2} \\ &=\frac{1+2\cos^2(x)\sec^2(x)2}{\sec^2(x)2} \\ &=\frac{2\cos^2(x)1\sec^2(x)}{\sec^2(x)2} \\ &=\frac{\cos(2x)\sec^2(x)}{\sec^2(x)2} \\ &=\frac{\cos^2(x)\cos(2x)1}{12\cos^2(x)} \\ &=\cos^2(x)+\sec(2x)\end{align*}$$ 
January 6th, 2018, 03:08 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
$$\begin{align*}\sin^2(x)\frac{\tan^2(x)+3}{1\tan^2(x)}&=\sin^2(x)\frac{2\tan(x)}{1\tan^2(x)}\frac{\tan^2(x)+3}{2\tan(x)} \\ &=\sin^2(x)\tan(2x)(\tan(x)+3\cot(x))/2 \\ &=\sin^2(x)\sin(2x)\sec(2x)(\tan(x)+3\cot(x))/2 \\ &=\sin^2(x)\sin(x)\cos(x)\sec(2x)(\tan(x)+3\cot(x) ) \\ &=\sin^2(x)\sec(2x)(\sin^2(x)+3\cos^2(x)) \\ &=\sin^2(x)\sec(2x)(1+2\cos^2(x)) \\ &=(1\cos^2(x))\sec(2x)(1+2\cos^2(x)) \\ &=(1\cos^2(x))(\sec(2x)+2\cos^2(x)\sec(2x)) \\ &=\sec(2x)+2\cos^2(x)\sec(2x)\cos^2(x)\sec(2x)2\cos^4(x)\sec(2x) \\ &=\sec(2x)+\cos^2(x)\sec(2x)2\cos^4(x)\sec(2x) \\ &=\sec(2x)+\cos^2(x)\sec(2x)(12\cos^2(x)) \\ &=\sec(2x)\cos^2(x)\sec(2x)(2\cos^2(x)1) \\ &=\sec(2x)\cos^2(x)\sec(2x)\cos(2x) \\ &=\sec(2x)\cos^2(x)\end{align*}$$ 
January 7th, 2018, 02:35 PM  #6 
Senior Member Joined: Oct 2009 Posts: 406 Thanks: 141 
You can very easily and mechanically prove a LOT of trig statements using the tangent halfangle 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, 09:13 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
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, 01:54 AM  #8 
Senior Member Joined: Oct 2009 Posts: 406 Thanks: 141  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, 05:35 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
Of course, but "etc." is a way of avoiding being specific.


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