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December 18th, 2017, 08:50 AM   #1
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Prove that : sin^2(x)*(tan^2(x)+3)/(1-tan^2(x))=sec(2x)-cos^2(x)
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December 18th, 2017, 10:12 AM   #2
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It's not an identity.
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December 18th, 2017, 03:57 PM   #3
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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)
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December 29th, 2017, 05:31 PM   #4
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$$\begin{align*}\sin^2(x)\frac{\tan^2(x)+3}{1-\tan^2(x)}&=-\sin^2(x)\frac{\sec^2(x)+2}{\sec^2(x)-2} \\
&=(\cos^2(x)-1)\frac{\sec^2(x)+2}{\sec^2(x)-2} \\
&=\frac{1+2\cos^2(x)-\sec^2(x)-2}{\sec^2(x)-2} \\
&=\frac{2\cos^2(x)-1-\sec^2(x)}{\sec^2(x)-2} \\
&=\frac{\cos(2x)-\sec^2(x)}{\sec^2(x)-2} \\
&=\frac{\cos^2(x)\cos(2x)-1}{1-2\cos^2(x)} \\
&=-\cos^2(x)+\sec(2x)\end{align*}$$

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January 6th, 2018, 03:08 AM   #5
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$$\begin{align*}\sin^2(x)\frac{\tan^2(x)+3}{1-\tan^2(x)}&=\sin^2(x)\frac{2\tan(x)}{1-\tan^2(x)}\frac{\tan^2(x)+3}{2\tan(x)} \\
&=\sin^2(x)\tan(2x)(\tan(x)+3\cot(x))/2 \\
&=\sin^2(x)\sin(2x)\sec(2x)(\tan(x)+3\cot(x))/2 \\
&=\sin^2(x)\sin(x)\cos(x)\sec(2x)(\tan(x)+3\cot(x) ) \\
&=\sin^2(x)\sec(2x)(\sin^2(x)+3\cos^2(x)) \\
&=\sin^2(x)\sec(2x)(1+2\cos^2(x)) \\
&=(1-\cos^2(x))\sec(2x)(1+2\cos^2(x)) \\
&=(1-\cos^2(x))(\sec(2x)+2\cos^2(x)\sec(2x)) \\
&=\sec(2x)+2\cos^2(x)\sec(2x)-\cos^2(x)\sec(2x)-2\cos^4(x)\sec(2x) \\
&=\sec(2x)+\cos^2(x)\sec(2x)-2\cos^4(x)\sec(2x) \\
&=\sec(2x)+\cos^2(x)\sec(2x)(1-2\cos^2(x)) \\
&=\sec(2x)-\cos^2(x)\sec(2x)(2\cos^2(x)-1) \\
&=\sec(2x)-\cos^2(x)\sec(2x)\cos(2x) \\
&=\sec(2x)-\cos^2(x)\end{align*}$$

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January 7th, 2018, 02:35 PM   #6
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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.
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January 7th, 2018, 09:13 PM   #7
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It doesn't "always work". For example, it wouldn't show that sin($\pi$/6 + x) + sin($\pi$/6 - x) ≡ cos(x).
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January 8th, 2018, 01:54 AM   #8
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Quote:
Originally Posted by skipjack View Post
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.
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January 8th, 2018, 05:35 AM   #9
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Of course, but "etc." is a way of avoiding being specific.
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