Results 1 to 7 of 7

Math Help - Trigonometry question.

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    72

    Trigonometry question.

    Hi all,

    I am new here by the way, I have a small problem for a question I can complete in Compound Angle identities

    Prove that:
    sinθ = tan (θ/2)(1 + cosθ)

    Look forward to the answer. I can almost work it work it out but I get stuck somewhere near the end.
    Last edited by mr fantastic; February 6th 2011 at 11:22 AM. Reason: Deleted begging in title.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    Hello and welcome to MathHelpForum.
    You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2011
    Posts
    72
    Quote Originally Posted by Plato View Post
    Hello and welcome to MathHelpForum.
    You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.
    I understand that. This is not homework, it's for my own understanding of Trig (revision). Okay how I did it...

    sinθ = tan (θ/2)(1 + cosθ)
    tan (θ/2) (1 + cos (θ/2 + θ/2))
    tan (θ/2)(cos^2(θ/2) - sin^2(θ/2))
    tan (θ/2)(cos^2(θ/2) + sin^2(θ/2) + cos^2(θ/2) - sin^2(θ/2))
    Trigonometry question.-help00.png
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Let A = \dfrac{\theta}{2} (purely to save me effort)

    From line 2: \tan(A)(1+\cos(2A)) = \dfrac{\sin(A)}{\cos(A)} \cdot 2\cos^2A = 2 \sin A \cos A = 2 \sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)

    There is a common identity you can use now to get the LHS
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2011
    Posts
    72
    Quote Originally Posted by e^(i*pi) View Post
    Let A = \dfrac{\theta}{2} (purely to save me effort)

    From line 2: \tan(A)(1+\cos(2A)) = \dfrac{\sin(A)}{\cos(A)} \cdot 2\cos^2A = 2 \sin A \cos A = 2 \sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)

    There is a common identity you can use now to get the LHS
    Hello e^(i*pi)

    I actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sinθ.
    Trigonometry question.-h1.png

    So this is correct?
    Trigonometry question.-he12.png

    Thanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,806
    Thanks
    697
    Hello, Googl!

    \text{Prove that: }\:\sin\theta \:=\:\tan \frac{\theta}{2}(1 + \cos\theta)

    Identities: . \begin{array}{ccccccc}\cos^2\!\frac{\theta}{2} &=& \dfrac{1+\cos\theta}{2} &\Rightarrow& 1 + \cos\theta &=& 2\cos^2\!\frac{\theta}{2} \\ \\[-3mm]<br />
\sin\theta &=& 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{array}



    \text{On the right side we have:}

    . . \tan\frac{\theta}{2}(1 + \cos\theta)\;=\;\dfrac{\sin\frac{\theta}{2}}{\cos\  frac{\theta}{2}}(2\cos^2\!\frac{\theta}{2}) \;=\;2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \;=\;\sin\theta

    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Googl View Post
    Hello e^(i*pi)

    I actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sinθ.
    Click image for larger version. 

Name:	h1.png 
Views:	25 
Size:	3.8 KB 
ID:	20700

    So this is correct?
    Click image for larger version. 

Name:	he12.png 
Views:	25 
Size:	5.2 KB 
ID:	20701

    Thanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.
    Yes, that's good when starting at the right.

    To start at the left and end up at the right...

    \displaystyle\ sin\theta=sin\left(\frac{\theta}{2}+\frac{\theta}{  2}\right)=2sin\frac{\theta}{2}cos\frac{\theta}{2}

    using sin2A=2sinAcosA

    which gets us to the half angle.
    To get tan, we need cos under the sine, so multiply by

    \displaystyle\ 1=\frac{cos\frac{\theta}{2}}{cos\frac{\theta}{2}}

    \displaystyle\ 2sin\frac{\theta}{2}cos\frac{\theta}{2}\left[\frac{cos\frac{\theta}{2}}{cos\frac{\theta}{2}}\ri  ght]=\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}\l  eft[2cos^2\frac{\theta}{2}\right]

    Now use

    cos^2A=\frac{1}{2}(1+cos2A)\Rightarrow\ 2cos^2A=1+cos2A

    to get

    \displaystyle\ sin\theta=tan\frac{\theta}{2}\left[1+cos\left(\frac{\theta}{2}+\frac{\theta}{2}\right  )\right]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Help with trigonometry question
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 28th 2010, 06:44 AM
  2. trigonometry question
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: October 4th 2009, 01:20 AM
  3. Trigonometry Question
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: August 12th 2009, 09:49 AM
  4. trigonometry question.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 19th 2007, 04:12 PM
  5. Trigonometry question
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: May 23rd 2006, 02:46 AM

Search Tags


/mathhelpforum @mathhelpforum