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 February 16th, 2017, 12:31 AM #1 Member   Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0 Double and Half Angle Formulas I'm really in a bad mood and need help solving these two problems. I had a terrible Pre-Calculus teacher and know almost nothing about Trigonometry. We're starting to move more into Trig in my Calculus class and I'm lost. This is a formula I want to learn more about and I've tried to familiarize myself with Trig Identities, but I'm lost. Please someone show me a step by step way to solve these. I swear it's not homework, but I do have class in about 10 hours and I need to know these things before-hand. Thank you "Simplify the expression by using a Double-Angle Formula or a Half-Angle Formula."  Sin(4) ----------- 1+Cos(4)  1-Cos(6)theta ----------------- Sin(6)theta February 16th, 2017, 01:00 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 650 Math Focus: Yet to find out. Do you know what the double and half angle formulas are? Also just to confirm,  $\dfrac{\sin (4 \theta)}{1 + \cos(4 \theta)}$  $\dfrac{1 - \cos(6 \theta)}{\sin(6 \theta)}$ are the questions? Last edited by skipjack; February 16th, 2017 at 05:44 AM. February 16th, 2017, 03:41 AM   #3
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 Originally Posted by nbg273 I'm really in a bad mood and need help solving these two problems. I had a terrible Pre-Calculus teacher and know almost nothing about Trigonometry. We're starting to move more into Trig in my Calculus class and I'm lost. This is a formula I want to learn more about and I've tried to familiarize myself with Trig Identities, but I'm lost. Please someone show me a step by step way to solve these. I swear it's not homework, but I do have class in about 10 hours and I need to know these things before-hand. Thank you "Simplify the expression by using a Double-Angle Formula or a Half-Angle Formula."  Sin(4) ----------- 1+Cos(4)  1-Cos(6)theta ----------------- Sin(6)theta
\displaystyle \begin{align*} \frac{\sin{\left( 4\,x \right) }}{1 + \cos{(4\,x)}} &\equiv \frac{2\,\sin{\left( 2\,x \right)}\cos{\left( 2\,x \right) }}{1 + 1 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2\,\left[ 1 - \sin^2{(2\,x)} \right] } \\ &\equiv \frac{\sin{(2\,x)}\cos{(2\,x)}}{1 - \sin^2{(2\,x)}} \\ &\equiv \frac{\sin{(2\,x)}\cos{(2\,x)}}{\cos^2{(2\,x)}} \\ &\equiv \frac{\sin{(2\,x)}}{\cos{(2\,x)}} \\ &\equiv \tan{(2\,x)} \end{align*}

Can you simplify this any more? February 16th, 2017, 06:01 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,939 Thanks: 2210  Using cos(2A) ≡ 1 - 2sin²(A), sin(2A) ≡ 2sin(A)cos(A) and sin(A)/cos(A) = tan(A), $\dfrac{1 - \cos(6\theta)}{\sin(6\theta)} = \dfrac{2\sin^2(3\theta)} {2\sin(3\theta)\cos(3\theta)} = \dfrac{\sin(3\theta)}{\cos(3\theta)} = \tan(3\theta)$. February 16th, 2017, 07:56 AM   #5
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 Originally Posted by Joppy Do you know what the double and half angle formulas are? Also just to confirm,  $\dfrac{\sin (4 \theta)}{1 + \cos(4 \theta)}$  $\dfrac{1 - \cos(6 \theta)}{\sin(6 \theta)}$ are the questions?
For the first one, it's Sin 4(degree sign)/1+Cos 4(degree sign), but the second problem is correct.

Last edited by nbg273; February 16th, 2017 at 08:06 AM. February 16th, 2017, 08:03 AM   #6
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 Originally Posted by Prove It \displaystyle \begin{align*} \frac{\sin{\left( 4\,x \right) }}{1 + \cos{(4\,x)}} &\equiv \frac{2\,\sin{\left( 2\,x \right)}\cos{\left( 2\,x \right) }}{1 + 1 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2\,\left[ 1 - \sin^2{(2\,x)} \right] } \\ &\equiv \frac{\sin{(2\,x)}\cos{(2\,x)}}{1 - \sin^2{(2\,x)}} \\ &\equiv \frac{\sin{(2\,x)}\cos{(2\,x)}}{\cos^2{(2\,x)}} \\ &\equiv \frac{\sin{(2\,x)}}{\cos{(2\,x)}} \\ &\equiv \tan{(2\,x)} \end{align*} Can you simplify this any more?
Sorry for the confusion, but there should be no x in the problem.

For the 4s, it should be degrees, not x's. unless that's the same thing... ? February 16th, 2017, 09:33 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,939 Thanks: 2210 Obviously, $x$ can be 1$^\circ\!$. February 16th, 2017, 09:42 AM   #8
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 Originally Posted by skipjack Obviously, $x$ can be 1$^\circ\!$.
Oh, ok thanks. I don't know this stuff very well. February 16th, 2017, 09:42 AM #9 Member   Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0 Thank you, everyone. This helped a TON!  Tags angle, double, formulas, half 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 swm06 Algebra 5 May 10th, 2012 08:50 PM suomik1988 Algebra 1 November 29th, 2009 08:13 PM peteglenn Algebra 1 September 16th, 2009 03:41 PM mikeportnoy Algebra 5 November 14th, 2008 02:18 AM Ciqo Algebra 1 June 9th, 2008 12:31 PM

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