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Math Help - Compound Angles or Double Angles?

  1. #1
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    Cool Compound Angles or Double Angles?

    Hi, I am stuck these types of questions and I just can't make any sense of them.
    Do I apply compound angles or double angles or the simple trigno functions?

    Kindly show me a few steps and I will take it from there.

    Q) Find all the angles between 0 and 360 which satisfy the following equations:

    1) 2+sinycosy=2sin^2y

    2) 2tan2x+sec40=1

    3) 1+2sin(3z/2+75)=0



    Another question is that when they say sin40 or sin10+40, the 40,10 and 40 are all angles right, even if they don't write the degree sign on them?

    Thanks!
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  2. #2
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    They should be written with the degree sign, unless the angles are measured in radians, but in that case you will nearly always be given multiples of \pi.

    3) Assuming the angles are measured in degrees...

    1 + 2\sin{\left(\frac{3z}{2}+ 75 \right)^{\circ}} = 0

    2\sin{\left(\frac{3z}{2} + 75\right)^{\circ}}= -1

    \sin{\left(\frac{3z}{2} + 75\right)^{\circ}} = -\frac{1}{2}

    \left(\frac{3z}{2}\right)^{\circ} + 75^{\circ} = \left\{180^{\circ} + 30^{\circ}, 360^{\circ} - 30^{\circ}\right\} + 360^{\circ} n where n \in \mathbf{Z}

    \left(\frac{3z}{2}\right)^{\circ} + 75^{\circ} = \left\{ 210^{\circ}, 330^{\circ} \right\} + 360^{\circ}n

    \left(\frac{3z}{2}\right)^{\circ} = \left\{ 135^{\circ}, 255^{\circ}\right\} + 360^{\circ}n

    (3z)^{\circ} = \left\{270^{\circ}, 510^{\circ} \right\} + 720^{\circ}n

    z^{\circ} = \left\{90^{\circ}, 170^{\circ} \right\} + 240^{\circ}n.
    Last edited by Prove It; July 24th 2010 at 09:32 PM.
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  3. #3
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    Thanks, yes these are all in degrees :/
    Can someone help me with the other 2 now

    Thanks!
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    Quote Originally Posted by Prove It View Post
    z^{\circ} = \left\{90^{\circ}, 170^{\circ} \right\} + 240^{\circ}.
    Well I think you made a mistake somewhere because I am getting the following angles: 90, 170 and 330

    Thanks anyways
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    Quote Originally Posted by unstopabl3 View Post
    Hi, I am stuck these types of questions and I just can't make any sense of them.
    Do I apply compound angles or double angles or the simple trigno functions?

    Kindly show me a few steps and I will take it from there.

    Q) Find all the angles between 0 and 360 which satisfy the following equations:

    1) 2+sinycosy=2sin^2y


    Thanks!
    1=sin^2y+cos^2y\ \Rightarrow\ 2=2\left(sin^2y+cos^2y\right)

    2\left(sin^2y+cos^2y\right)+sinycosy=2sin^2y

    2\left(cos^2y+sin^2y+0.5sinycosy)=2sin^2y

    sin^2y=cos^2y+sin^2y+0.5sinycosy

    cos^2y+cosy(0.5siny)=0

    cosy(cosy+0.5siny)=0

    gives 2 solutions.

    Finish by writing cosy=-0.5siny\ \Rightarrow\ -2=tany
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  6. #6
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    Quote Originally Posted by unstopabl3 View Post
    Well I think you made a mistake somewhere because I am getting the following angles: 90, 170 and 330

    Thanks anyways
    \frac{3z}{2}+75^o=210^o,\ 330^o,\ 570^o,\ 690^o....

    \frac{3z}{2}=135^o,\ 255^o,\ 495^o,\ 615^o....

    3z=270^o,\ 510^o,\ 990^o,\ 1230^o....

    z=90^o,\ 170^o,\ 330^o.......


    (2) 2tan2x+sec40^o=1

    2tan2x+\frac{1}{cos40^o}=1

    2tan2x=1-\frac{1}{cos40^o}

    tan2x=\frac{cos40^o-1}{2cos40^0}

    That can be evaluated and you can finish with the identity tan2x=\frac{2tanx}{1-tan^2x}

    to get a quadratic equation in tanx......

    if you want some practice,
    though that identity is used to discover tan2x from tanx.

    Instead just take the inverse tan of both sides to get 2x between 0 and 720 degrees.
    Since tan(angle) is the slope of a line, remember there is a second solution 180 degrees away
    from the solution returned by a calculator.
    Last edited by Archie Meade; July 24th 2010 at 12:52 PM.
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    \frac{3z}{2}+75^o=210^o,\ 330^o,\ 570^o,\ 690^o....

    \frac{3z}{2}=135^o,\ 255^o,\ 495^o,\ 615^o....

    3z=270^o,\ 510^o,\ 990^o,\ 1230^o....
    z=90^o,\ 170^o,\ 330^o.......

    (2) 2tan2x+sec40^o=1

    2tan2x+\frac{1}{cos40^o}=1

    2tan2x=1-\frac{1}{cos40^o}

    tan2x=\frac{cos40^o-1}{2cos40^0}

    That can be evaluated and you can finish with the identity tan2x=\frac{2tanx}{1-tan^2x}

    to get a quadratic equation in tanx.
    Using that identity makes it complicated!
    Instead, it can be solved using the general solution for tanx. That's let tan2x=a => 2x= k.pi+Arctg(a) => x= ./5(k.pi+Arctg(a)) for evrey k in Z.
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  8. #8
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    Quote Originally Posted by unstopabl3 View Post
    Well I think you made a mistake somewhere because I am getting the following angles: 90, 170 and 330

    Thanks anyways
    I did not make a mistake.

    You keep adding multiples of 240^{\circ} until you have all possible solutions in your given domain.

    What is 90^{\circ} + 240^{\circ}?
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  9. #9
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    Quote Originally Posted by Prove It View Post
    I did not make a mistake.

    You keep adding multiples of 240^{\circ} until you have all possible solutions in your given domain.

    What is 90^{\circ} + 240^{\circ}?
    Well 240 didn't have a "n" written infront of it in your first attempt. Maybe that's why I was confused.
    Thanks for fixing that up
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