Results 1 to 4 of 4

Thread: trigonometry (1)

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    trigonometry (1)

    Without using mathematical tables or calculator , show that
    $\displaystyle \theta=\frac{\pi}{10}$ satisfies the equation
    $\displaystyle \sin2\theta=\cos3\theta $.



    Thanks for ur help .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Consider : $\displaystyle \cos^2 3\theta-\sin^2 2\theta$
    ------------------------------------------------------
    Remember these identities :
    $\displaystyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

    $\displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

    Multiply them :

    $\displaystyle \begin{aligned}
    \cos(a-b)\cos(a+b) &=\cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\
    &=\cos^2(b)(1-\sin^2(a))-\sin^2(a)\sin^2(b) \\
    &=\cos^2(b)-\sin^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\
    &=\cos^2(b)-\sin^2(a)(\cos^2(b)+\sin^2(b)) \\
    &=\boxed{\cos^2(b)-\sin^2(a)}
    \end{aligned}
    $
    ------------------------------------------------------

    Hence $\displaystyle \cos^2 3\theta-\sin^2 2\theta=\cos \theta \cos 5\theta$

    And if $\displaystyle \theta=\frac{\pi}{10}$, then $\displaystyle \cos 5\theta=0$

    So $\displaystyle 0=\cos^2 3\theta-\sin^2 2\theta=(\cos 3\theta-\sin 2\theta)(\cos 3\theta+\sin 2\theta)$ (1)

    Now notice that :
    $\displaystyle 3\theta=\frac{3\pi}{10}$ and $\displaystyle 2\theta=\frac\pi 5$ are both in the first quadrant (because $\displaystyle <\frac\pi 2$)
    So $\displaystyle \cos 3\theta>0$ and $\displaystyle \sin 2\theta>0$

    Hence $\displaystyle \cos 3\theta+\sin 2\theta>0$


    And it follows from (1) that $\displaystyle \boxed{\cos 3\theta-\sin 2\theta=0}$ if $\displaystyle \theta=\frac{\pi}{10}$


    Looks clear to you ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by thereddevils View Post
    Without using mathematical tables or calculator , show that

    $\displaystyle \theta=\frac{\pi}{10}$ satisfies the equation
    $\displaystyle \sin2\theta=\cos3\theta $.




    Thanks for ur help .
    Since it's a 'show question' (rather than a 'solve question') the following is perfectly acceptable:

    $\displaystyle \cos \left( \frac{3 \pi}{10} \right) = \sin \left(\frac{\pi}{2} - \frac{3 \pi}{10} \right) = \sin \left( \frac{2 \pi}{10} \right)$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2009
    Posts
    381
    Quote Originally Posted by Moo View Post
    Hello,

    Consider : $\displaystyle \cos^2 3\theta-\sin^2 2\theta$
    ------------------------------------------------------
    Remember these identities :
    $\displaystyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

    $\displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

    Multiply them :

    $\displaystyle \begin{aligned}
    \cos(a-b)\cos(a+b) &=\cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\
    &=\cos^2(b)(1-\sin^2(a))-\sin^2(a)\sin^2(b) \\
    &=\cos^2(b)-\sin^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\
    &=\cos^2(b)-\sin^2(a)(\cos^2(b)+\sin^2(b)) \\
    &=\boxed{\cos^2(b)-\sin^2(a)}
    \end{aligned}
    $
    ------------------------------------------------------

    Hence $\displaystyle \cos^2 3\theta-\sin^2 2\theta=\cos \theta \cos 5\theta$

    And if $\displaystyle \theta=\frac{\pi}{10}$, then $\displaystyle \cos 5\theta=0$

    So $\displaystyle 0=\cos^2 3\theta-\sin^2 2\theta=(\cos 3\theta-\sin 2\theta)(\cos 3\theta+\sin 2\theta)$ (1)

    Now notice that :
    $\displaystyle 3\theta=\frac{3\pi}{10}$ and $\displaystyle 2\theta=\frac\pi 5$ are both in the first quadrant (because $\displaystyle <\frac\pi 2$)
    So $\displaystyle \cos 3\theta>0$ and $\displaystyle \sin 2\theta>0$

    Hence $\displaystyle \cos 3\theta+\sin 2\theta>0$


    And it follows from (1) that $\displaystyle \boxed{\cos 3\theta-\sin 2\theta=0}$ if $\displaystyle \theta=\frac{\pi}{10}$


    Looks clear to you ?

    What a beautiful solution ! Thanks Moo . Thanks Mr F too , for that simple and easy approach
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry to Memorize, and Trigonometry to Derive
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: Feb 23rd 2017, 09:35 AM
  2. How To Do Trigonometry For This...?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Jan 10th 2009, 05:56 PM
  3. How To Do This Trigonometry
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jan 3rd 2009, 01:55 AM
  4. trigonometry
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Dec 31st 2008, 08:06 PM
  5. Trigonometry
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Dec 18th 2008, 04:40 PM

Search Tags


/mathhelpforum @mathhelpforum