Results 1 to 5 of 5

Thread: Angle Question

  1. #1
    Newbie
    Joined
    Jan 2006
    Posts
    2

    Angle Question

    So I have two angle problems...

    sin2x = 2cosxcos2x ; which becomes
    0 = 2cosxcos2x - sin2x
    0 = 2cosxcos2x - 2sinxcosx
    0 = 2cosx(cos2x - sin x)

    so either

    2cosx = 0 OR cos2x - sinx = 0

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CrimesofParis
    So I have two angle problems...

    sin2x = 2cosxcos2x ; which becomes
    0 = 2cosxcos2x - sin2x
    0 = 2cosxcos2x - 2sinxcosx
    0 = 2cosx(cos2x - sin x)

    so either

    2cosx = 0 OR cos2x - sinx = 0

    Thank you.

    Okay you did everything correctly up till now. Now you are required to solved for $\displaystyle x$.
    The first one $\displaystyle 2\cos x=0$. Divide by two,
    $\displaystyle \cos x=0$, now you need to know when $\displaystyle \cos x$ is zero. That is when $\displaystyle x=90,270$ as two non-coterminal angels. To find all solutions it is the same as $\displaystyle x=90+360k$ and $\displaystyle x=270+360k=-90+360+360k=-90+360(1+k)$ where $\displaystyle k$ is any integer. Because as you know that addition and subtraction of 360 keeps the trigonometric function of that angel the same. But $\displaystyle k+1$ is also an integer. Thus, all solutions for $\displaystyle x$ have form. $\displaystyle x=(-1)^{k+1}90+360k, k \epsilon Z $ or in radian form $\displaystyle x=(-1)^{k+1}\frac{\pi}{2}+2\pi k,k \epsilon Z$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    For your second question:
    $\displaystyle \cos 2x-\sin x=0$
    Use the identity
    $\displaystyle 1-2\sin^2x=\cos 2x$
    Thus,
    $\displaystyle 1-2\sin^2x-\sin x=0$
    Change signs,
    $\displaystyle 2\sin^2x+\sin x-1=0$
    Factor,
    $\displaystyle (2\sin x-1)(\sin x +1)=0$
    Now each one is set to zero,
    $\displaystyle 2\sin x-1=0$
    Thus,
    $\displaystyle \sin x=\frac{1}{2}$
    That is when $\displaystyle x=30,150$ as the only two non-coterminal angels. Thus all solution have form $\displaystyle x=30+360k, 150+360k$ But,
    $\displaystyle x=30+360k=30+180(2k)$
    $\displaystyle x=150+360k=-30+180(2k+1)$
    Thus, all solutions have form $\displaystyle x=(-1)^k30+180k$ or in radian form $\displaystyle x=(-1)^k\frac{\pi}{6}+\pi k,k \epsilon Z$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Finally the last factor equal to zero, (of the quadratic)
    $\displaystyle \sin x+1=0$
    Thus,
    $\displaystyle \sin x=-1$ That happens when $\displaystyle x=270$ and that is the only non-coterminal angle. Thus all solutions have form $\displaystyle x=270+360k$. But $\displaystyle 270+360k=-90+360+360k=-90+360(k+1)$.
    Thus, all solutions are $\displaystyle x=-90+360k$. Or in radian form $\displaystyle x=-\frac{\pi}{2}+2\pi k,k\epsilon Z$.

    If you have difficulty understanding all that manipulating after the solutions just understand that by adding 360 or subtracting 360 you change nothing. Thus, those are all the solutions. All I did was manipulated into a more elegant form.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2006
    Posts
    2
    Thank you so much, that pretty much answers my questions, except I forgot to mention the answer has to be radians, so I should be able to covert the middle steps to radians. But thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double angle question help
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Apr 29th 2011, 12:30 PM
  2. Replies: 4
    Last Post: Oct 15th 2010, 08:56 AM
  3. Need help with this Angle of Depression question
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 4th 2009, 04:07 AM
  4. Replies: 2
    Last Post: Mar 28th 2009, 02:18 PM
  5. Double Angle Question
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Mar 7th 2009, 12:09 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum