My Math Forum Double and Half Angle Formulas

 Trigonometry Trigonometry Math Forum

 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." [1] Sin(4) ----------- 1+Cos(4) [2] 1-Cos(6)theta ----------------- Sin(6)theta
 February 16th, 2017, 01:00 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,798 Thanks: 634 Math Focus: Yet to find out. Do you know what the double and half angle formulas are? Also just to confirm, [1] $\dfrac{\sin (4 \theta)}{1 + \cos(4 \theta)}$ [2] $\dfrac{1 - \cos(6 \theta)}{\sin(6 \theta)}$ 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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Quote:
 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." [1] Sin(4) ----------- 1+Cos(4) [2] 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: 20,622 Thanks: 2076 [2] 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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Quote:
 Originally Posted by Joppy Do you know what the double and half angle formulas are? Also just to confirm, [1] $\dfrac{\sin (4 \theta)}{1 + \cos(4 \theta)}$ [2] $\dfrac{1 - \cos(6 \theta)}{\sin(6 \theta)}$ 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: 20,622 Thanks: 2076 Obviously, $x$ can be 1$^\circ\!$.
February 16th, 2017, 09:42 AM   #8
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Quote:
 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!

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