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Math Help - Trignometric Identity Questions

  1. #1
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    Trignometric Identity Questions

    I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

    If somebody could help me with these questions that would be great.

    1. Simplify:
    cos2A/(cosA+sinA)

    2. Prove:
    sin^2(2)A=sin^2A-sin^4A

    3. sqrt(2)cos(A-(pi/4))=cosA+sinA

    4. Solve for all possible values of X:
    -5sin(15(x-17degrees))-3=0

    excuse my lack of math typing skills
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  2. #2
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    Quote Originally Posted by Liam View Post
    I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

    If somebody could help me with these questions that would be great.

    1. Simplify:
    cos2A/(cosA+sinA)
    \frac{\cos 2A}{\cos A + \sin A}

    =\frac{\cos^2 A - \sin^2 A}{\cos A + \sin A}

    =\frac{(\cos A + \sin A)(\cos A - \sin A)}{\cos A + \sin A}

    =\cos A - \sin A

    we used the formula here, \cos 2A=\cos^2 A - \sin^2 A

    and, a^2 - b^2 = (a+b)(a-b)
    Last edited by Shyam; November 8th 2008 at 06:28 PM.
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  3. #3
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    Quote Originally Posted by Liam View Post
    I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

    If somebody could help me with these questions that would be great.
    2. Prove:
    sin^2(2)A=sin^2A-sin^4A
    Right Side = \sin^2 A(1 - \sin^2 A)

    = \sin^2 A \cos^2 A

    = \frac{4\sin^2 A \cos^2 A}{4}

    = \frac{(2\sin A \cos A)^2}{4}

    = \frac{\sin^2 2A}{4}

    we used the formula here, 2 \sin A \cos A = \sin 2A
    Last edited by Shyam; November 8th 2008 at 06:23 PM.
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  4. #4
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    Quote Originally Posted by Liam View Post
    I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

    If somebody could help me with these questions that would be great.
    3. sqrt(2)cos(A-(pi/4))=cosA+sinA
    =\sqrt{2} \cos \left(A - \frac{\pi}{4}\right)

    = \sqrt{2}\left[\cos A .\cos \left(\frac{\pi}{4}\right) + \sin A .\sin \left(\frac{\pi}{4}\right)\right]

    = \sqrt{2} \left( \cos A .\frac{1}{\sqrt{2}}+\sin A. \frac{1}{\sqrt{2}}\right)

    = \cos A + \sin A

    Formula used : \cos \left(\alpha - \beta\right) = \cos \alpha .\cos \beta + \sin \alpha .\sin \beta
    Last edited by Shyam; November 8th 2008 at 06:33 PM.
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  5. #5
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    Quote Originally Posted by Liam View Post
    I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

    If somebody could help me with these questions that would be great.

    4. Solve for all possible values of X:
    -5sin(15(x-17degrees))-3=0

    excuse my lack of math typing skills
    - 5\sin \left[ {15\left( {x - 17^\circ } \right)} \right] - 3 = 0 \hfill \\

     - 5\sin \left[ {15\left( {x - 17^\circ } \right)} \right] = 3 \hfill \\

    \sin \left[ {15\left( {x - 17^\circ } \right)} \right] = \frac{3}<br />
{{ - 5}} \hfill \\

    15\left( {x - 17^\circ } \right) = \sin ^{ - 1} \left( { - \frac{3}<br />
{5}} \right) \hfill \\

    {\text{Now, }}\sin ^{ - 1} \left( { - \frac{3}<br />
{5}} \right) =  - 36.87^\circ  =  - 37^\circ  \hfill \\

    \sin \theta {\text{ is negative in 3rd and 4th quadrants, so, angle is}} \hfill \\

    = 180^\circ  + 37^\circ {\text{   and    }} = 360^\circ  - 37^\circ  \hfill \\

    = 217^\circ ,323^\circ  \hfill \\

    15\left( {x - 17^\circ } \right) = 217^\circ  \hfill \\

    x - 17^\circ  = 14.5^\circ  \hfill \\

    x = 14.5^\circ  + 17^\circ  = 31.5^\circ  \hfill \\

    {\text{Also,}} \hfill \\

    15\left( {x - 17^\circ } \right) = 323^\circ  \hfill \\

    {\text{Now, finish it, in the same way}}{\text{.}} \hfill \\
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  6. #6
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    Thanks Shyam, you were of great help!
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