My Math Forum  

Go Back   My Math Forum > High School Math Forum > Trigonometry

Trigonometry Trigonometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
February 16th, 2017, 01:31 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
nbg273 is offline  
 
February 16th, 2017, 02:00 AM   #2
Senior Member
 
Joined: Feb 2016
From: Australia

Posts: 1,397
Thanks: 479

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.
Joppy is offline  
February 16th, 2017, 04:41 AM   #3
Member
 
Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

Quote:
Originally Posted by nbg273 View Post
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?
Prove It is offline  
February 16th, 2017, 07:01 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 18,142
Thanks: 1417

[2] Using cos(2A) ≡ 1 - 2sin²(A), sin(2A) ≡ 2sin(A)cos(A) and sin(A)/cos(A) = tan(A),

$\dfrac{1 - \cos(6\theta)}{\sin(6\theta)} = \dfrac{2\sin^2(3\theta)} {2\sin(3\theta)\cos(3\theta)} = \dfrac{\sin(3\theta)}{\cos(3\theta)} = \tan(3\theta)$.
skipjack is offline  
February 16th, 2017, 08:56 AM   #5
Member
 
Joined: Feb 2017
From: East U.S.

Posts: 40
Thanks: 0

Quote:
Originally Posted by Joppy View Post
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 09:06 AM.
nbg273 is offline  
February 16th, 2017, 09:03 AM   #6
Member
 
Joined: Feb 2017
From: East U.S.

Posts: 40
Thanks: 0

Quote:
Originally Posted by Prove It View Post
$\displaystyle \begin{align*} \frac{\sin{\left( 4\,x \right) }}{1 + \cos{(4\,x)}} &\equiv \frac{2\,\sin{\left( 2\,x \right)}\cos{\left( 2\,x \right) }}{1 + 1 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2 - 2\,\sin^2{(2\,x)}} \\ &\equiv \frac{2\,\sin{(2\,x)}\cos{(2\,x)}}{2\,\left[ 1 - \sin^2{(2\,x)} \right] } \\ &\equiv \frac{\sin{(2\,x)}\cos{(2\,x)}}{1 - \sin^2{(2\,x)}} \\ &\equiv \frac{\sin{(2\,x)}\cos{(2\,x)}}{\cos^2{(2\,x)}} \\ &\equiv \frac{\sin{(2\,x)}}{\cos{(2\,x)}} \\ &\equiv \tan{(2\,x)} \end{align*}$

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... ?
nbg273 is offline  
February 16th, 2017, 10:33 AM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 18,142
Thanks: 1417

Obviously, $x$ can be 1$^\circ\!$.
skipjack is offline  
February 16th, 2017, 10:42 AM   #8
Member
 
Joined: Feb 2017
From: East U.S.

Posts: 40
Thanks: 0

Quote:
Originally Posted by skipjack View Post
Obviously, $x$ can be 1$^\circ\!$.
Oh, ok thanks. I don't know this stuff very well.
nbg273 is offline  
February 16th, 2017, 10:42 AM   #9
Member
 
Joined: Feb 2017
From: East U.S.

Posts: 40
Thanks: 0

Thank you, everyone. This helped a TON!
nbg273 is offline  
Reply

  My Math Forum > High School Math Forum > Trigonometry

Tags
angle, double, formulas, half



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Half Angle Identity swm06 Algebra 5 May 10th, 2012 09:50 PM
what half angle identity would I use? suomik1988 Algebra 1 November 29th, 2009 09:13 PM
Multiple Choice Problem on Using Double-Angle Formulas peteglenn Algebra 1 September 16th, 2009 04:41 PM
half angle .. mikeportnoy Algebra 5 November 14th, 2008 03:18 AM
Half-Angle Formulas Ciqo Algebra 1 June 9th, 2008 01:31 PM





Copyright © 2017 My Math Forum. All rights reserved.