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Math Help - A few problems...Compound & double-angle trig. formulae

  1. #1
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    A few problems...Compound & double-angle trig. formulae

    Im stuck on a few questions...

    Any ideas?

    1) Solve the following equation for θ in the interval 0 ≤ θ ≤ 2π.

    sin( 2 θ) +5/9 cos( θ) = 0


    2) Solve
    4 cos( 2 θ) = 1+tan( θ)

    for θ in the interval −π/2 < θ < π/2.



    3) Which of the following is equal to

    (sin θ + cos θ)^2 + (sin θ − cos θ)^2 for all θ?

    A) 2 cos(2 θ)
    B) 2
    C) 2 sin^2θ
    D) cos(2 θ) − sin(2 θ)
    E) 2 sin(2 θ)
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  2. #2
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    1) sin( 2*beta ) + 5/9 * cos( beta ) = 0

    sin( 2*beta ) = 2*sin( beta )*cos( beta )

    sin( 2*beta ) + 5/9 * cos( beta ) = 2*sin( beta )*cos( beta ) + 5/9 * cos( beta ) = 0

    cos( beta )*[ 2*sin( beta ) + 5/9 ] = 0

    a) cos( beta ) = 0 => beta = pi/2 or beta = 3*pi/2

    b) 2*sin( beta ) + 5/9 = 0

    sin( beta ) = -5/18 => beta = arcsin( -5/18 ) ..... use calculator

    <-------------------------------------->

    2) it is more complicated.... probably will solve it later

    // i ' ve tried to solve it but it is complcated so the answer is very long and i didnt finished it, probably there is some faster way , some kind of trick...

    <--------------------------------------->

    3)
    the right answer is B) 2

    ( sin( beta ) + cos( beta ) )^2 + ( sin( beta ) - cos( beta ) )^2 =
    = [ sin^2( beta ) + 2*sin( beta )*cos( beta ) + cos^2( beta ) ] + [ sin^2( beta ) - 2*sin( beta )*cos( beta ) + cos^2( beta ) ] =
    = 2*( sin^2( beta ) + cos^2( beta ) ) = 2*1 = 2
    Last edited by josipive; May 13th 2009 at 03:03 PM.
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  3. #3
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    Quote Originally Posted by josipive View Post
    1) sin( 2*beta ) + 5/9 * cos( beta ) = 0

    sin( 2*beta ) = 2*sin( beta )*cos( beta )

    sin( 2*beta ) + 5/9 * cos( beta ) = 2*sin( beta )*cos( beta ) + 5/9 * cos( beta ) = 0

    cos( beta )*[ 2*sin( beta ) + 5/9 ] = 0

    a) cos( beta ) = 0 => beta = pi/2 or beta = 3*pi/2

    b) 2*sin( beta ) + 5/9 = 0

    sin( beta ) = -5/18 beta = arcsin( -5/18 ) ..... use calculator
    Shouldnt there be 4 answers...or I could be way off...
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  4. #4
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    Quote Originally Posted by Maccabhoy View Post
    Shouldnt there be 4 answers...or I could be way off...
    2 different answers for cos( beta ) = 0
    and 2 two different answers for arcsin( -5/18 )

    for arcsin( -5/18 ) calculator will give you answer between -pi/2 and pi/2
    but you have one more answer and that is arcsin( -5/18 ) - pi/2 ( I think )

    also look at the first post to see answer for 3)

    // not arcsin( -5/18 ) - pi/2 , but - pi - arcsin( -5/18 )
    Last edited by josipive; May 13th 2009 at 03:28 PM.
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  5. #5
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    Thanks josipive.

    The explanations are great!
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  6. #6
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    Hello, Maccabhoy!

    Here's the last one . . .



    3) Which of the following is equal to . (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 .for all \theta ?

    . . (A)\;2\cos(2\theta) \qquad(B)\;2 \qquad <br />
(C)\;2\sin^2\!\theta \qquad (D)\;\cos(2\theta) - \sin(2\theta) \qquad(E)\;2\sin(2\theta)
    Did you even try to multiply it out ??


    (\sin\theta + \cos\theta)^2 + (\sin\theta -\cos\theta)^2

    . . = \;\sin^2\!\theta + 2\sin\theta\cos\theta + \cos^2\!\theta + \sin^2\!\theta - 2\sin\theta\cos\theta + \cos^2\!\theta

    . . = \;2\sin^2\!\theta + 2\cos^2\!\theta \;=\;2\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \;=\;2(1) \;=\;2

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  7. #7
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    Quote Originally Posted by josipive View Post
    2 different answers for cos( beta ) = 0
    and 2 two different answers for arcsin( -5/18 )

    for arcsin( -5/18 ) calculator will give you answer between -pi/2 and pi/2
    but you have one more answer and that is arcsin( -5/18 ) - pi/2 ( I think )

    also look at the first post to see answer for 3)

    // not arcsin( -5/18 ) - pi/2 , but - pi - arcsin( -5/18 )
    Hmm... arcsin (-5/18) and - pi - arcsin( -5/18 ) are giving me wring answers.

    I am getting -0.28 and -2.86 respectively for these, but the quiz says they are wrong answers...and yes my calculator is in rad...
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  8. #8
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    anyone know how figure this one out?

    2) Solve
    4 cos( 2 θ) = 1+tan( θ)

    for θ in the interval −π/2 < θ < π/2.
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  9. #9
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    Quote Originally Posted by Maccabhoy View Post
    Hmm... arcsin (-5/18) and - pi - arcsin( -5/18 ) are giving me wring answers.

    I am getting -0.28 and -2.86 respectively for these, but the quiz says they are wrong answers...and yes my calculator is in rad...
    dont know... on my calculator everything works fine

    1.angle ) arcsin( -5/18 )

    2.angle ) -pi - arcsin( -5/18 ) // try -pi + arcsin( 5/18 ) that is the same
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