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February 16th, 2017, 01:31 AM  #1 
Member Joined: Feb 2017 From: East U.S. Posts: 33 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 PreCalculus 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 beforehand. Thank you "Simplify the expression by using a DoubleAngle Formula or a HalfAngle Formula." [1] Sin(4)  1+Cos(4) [2] 1Cos(6)theta  Sin(6)theta 
February 16th, 2017, 02:00 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,225 Thanks: 420 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 06:44 AM. 
February 16th, 2017, 04:41 AM  #3  
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Quote:
Can you simplify this any more?  
February 16th, 2017, 07:01 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312 
[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, 08:56 AM  #5 
Member Joined: Feb 2017 From: East U.S. Posts: 33 Thanks: 0  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 09:06 AM. 
February 16th, 2017, 09:03 AM  #6  
Member Joined: Feb 2017 From: East U.S. Posts: 33 Thanks: 0  Quote:
For the 4s, it should be degrees, not x's. unless that's the same thing... ?  
February 16th, 2017, 10:33 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312 
Obviously, $x$ can be 1$^\circ\!$.

February 16th, 2017, 10:42 AM  #8 
Member Joined: Feb 2017 From: East U.S. Posts: 33 Thanks: 0  
February 16th, 2017, 10:42 AM  #9 
Member Joined: Feb 2017 From: East U.S. Posts: 33 Thanks: 0 
Thank you, everyone. This helped a TON! 

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angle, double, formulas, half 
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