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Math Help - exact value using trig identities

  1. #1
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    Arrow exact value using trig identities

    please help with these problems thanks!

    1.
    cos^2(75degrees)
    2.
    sec^2(15degrees)


    also
    FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
    sin(x-B)

    I got (2(squareroot)6-1)/6 but Im not 100% sure.


    Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.





    Any help is gratefully appreciated!!!
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  2. #2
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    Quote Originally Posted by amanda0603 View Post
    please help with these problems thanks!

    1.
    cos^2(75degrees)
    2.
    sec^2(15degrees)


    also
    FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
    sin(x-B)

    I got (2(squareroot)6-1)/6 but Im not 100% sure.


    Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.





    Any help is gratefully appreciated!!!
    \cos^2{75^\circ} = (\cos{75^\circ})^2

     = [\cos{(60^\circ + 15^\circ)}]^2.


    Now using the sum identity for cosine, \cos{(a + b)} = \cos{a}\cos{b} - \sin{a}\sin{b}, this becomes...


    (\cos{60^\circ}\cos{15^\circ} - \sin{60^\circ}\sin{15^\circ})^2

     = \left(\frac{1}{2}\cos{15^\circ} - \frac{\sqrt{3}}{2}\sin{15^\circ}\right)^2.


    Now you need the half angle identities, \sin{\frac{\theta}{2}} = \pm\sqrt{\frac{1 - \cos{\theta}}{2}} and \cos{\frac{\theta}{2}} = \pm\sqrt{\frac{1 + \cos{\theta}}{2}}. Notice that 15 = \frac{30}{2}.


    \left(\frac{1}{2}\cos{15^\circ} - \frac{\sqrt{3}}{2}\sin{15^\circ}\right)^2 = \left(\frac{1}{2}\sqrt{\frac{1 + \cos{30^\circ}}{2}} - \frac{\sqrt{3}}{2}\sqrt{\frac{1 - \cos{30^\circ}}{2}}\right)^2

     = \left(\frac{1}{2}\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} - \frac{\sqrt{3}}{2}\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}\right)^2.

    I'll leave you to clean that up...
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  3. #3
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    Quote Originally Posted by amanda0603 View Post
    please help with these problems thanks!

    1.
    cos^2(75degrees)
    2.
    sec^2(15degrees)


    also
    FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
    sin(x-B)

    I got (2(squareroot)6-1)/6 but Im not 100% sure.


    Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.





    Any help is gratefully appreciated!!!
    2. \sec^2{15^\circ} = \frac{1}{\cos^2{15^\circ}}

     = \frac{1}{(\cos{15^\circ})^2}

    Now use the half angle identity \cos{\frac{\theta}{2}} = \pm \sqrt{\frac{1 + \cos{\theta}}{2}}

     = \frac{1}{\left(\sqrt{\frac{1 + \cos{30^\circ}}{2}}\right)^2}

     = \frac{1}{\left(\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}\right)^2}

     = \frac{1}{\frac{1 + \frac{\sqrt{3}}{2}}{2}}

     = \frac{2}{1 + \frac{\sqrt{3}}{2}}


    (and if you want to rationalise the denominator...)


     = \frac{2}{\frac{2 + \sqrt{3}}{2}}

     = \frac{4}{2 + \sqrt{3}}

     = \frac{4(2 - \sqrt{3})}{4 - 3}

     = 8 - 4\sqrt{3}.
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  4. #4
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    Quote Originally Posted by amanda0603 View Post
    please help with these problems thanks!

    1.
    cos^2(75degrees)
    2.
    sec^2(15degrees)


    also
    FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
    sin(x-B)

    I got (2(squareroot)6-1)/6 but Im not 100% sure.


    Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.





    Any help is gratefully appreciated!!!
    3. \cos{x} = \frac{1}{3}

    You need to use the Pythagorean Identity, \sin^2{x} + \cos^2{x} = 1.

    \sin^2{x} + \cos^2{x} = 1

    \sin^2{x} + \left(\frac{1}{3}\right)^2 = 1

    \sin^2{x} + \frac{1}{9} = 1

    \sin^2{x} = \frac{8}{9}

    \sin{x} = \frac{\sqrt{8}}{3}

    \sin{x} = \frac{2\sqrt{2}}{3} (We know that \sin{x} > 0 since x is in the first quadrant).


    We also know \sin{B} = -\frac{1}{2}. Once again, use the Pythagorean Identity.

    \sin^2{B} + \cos^2{B} = 1

    \left(-\frac{1}{2}\right)^2 + \cos^2{B} = 1

    \frac{1}{4} + \cos^2{B} = 1

    \cos^2{B} = \frac{3}{4}

    \cos{B} = \frac{\sqrt{3}}{2} (we know that \cos{B} > 0 because B is in the fourth quadrant.


    Now we have \sin{x}, \cos{x}, \sin{B}, \cos{B}.

    We were trying to find \sin{(x - B)}.

    For this we use the difference formula...

    \sin{(x - B)} = \sin{x}\cos{B} - \cos{x}\sin{B}

     = \frac{2\sqrt{2}}{3}\left(\frac{\sqrt{3}}{2}\right) - \frac{1}{3}\left(-\frac{1}{2}\right)

     = \frac{\sqrt{6}}{3} + \frac{1}{6}

     = \frac{2\sqrt{6} + 1}{6}.


    So you were VERY close...
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    Quote Originally Posted by amanda0603 View Post
    please help with these problems thanks!

    Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.
    Find the exact value of what?
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  6. #6
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    IM SORRY!!!
    the exact value of tan2A


    THANKS SO MUCH!
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  7. #7
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    Quote Originally Posted by amanda0603 View Post
    please help with these problems thanks!

    Find the exact value of tan2A given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.
    You need to use the following identities...

    \sin^2{x} + \cos^2{x} = 1,

    \tan{x} = \frac{\sin{x}}{\cos{x}}

    \tan{2x} = \frac{2\tan{x}}{1 - \tan^2{x}}.


    First...

    \sin^2{A} + \cos^2{A} = 1

    \left(-\frac{4}{5}\right)^2 + \cos^2{A} = 1

    \frac{16}{25} + \cos^2{A} = 1

    \cos^2{A} = \frac{9}{25}

    \cos{A} = \frac{3}{5} (We know that \cos{A} > 0 because A is in quadrant 4.


    Now...

    \tan{A} = \frac{\sin{A}}{\cos{A}}

     = \frac{-\frac{4}{5}}{\frac{3}{5}}

     = -\frac{4}{3}.


    So

    \tan{2A} = \frac{2\tan{A}}{1 - \tan^2{A}}

     = \frac{2\left(-\frac{4}{3}\right)}{1 - \left(-\frac{4}{3}\right)^2}

     = \frac{-\frac{8}{3}}{1 - \frac{16}{9}}

     = \frac{-\frac{8}{3}}{-\frac{7}{9}}

     = \frac{24}{7}.
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