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Math Help - Help with this problems!asap

  1. #1
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    Help with this problems!asap

    New here so be nice

    How do you find the exact value of these expressions?

    sin^2 15(degrees) cot^2 75(degrees) + cos^2 15(degrees) tan^2 75(degrees)


    and


    1-tan(pi/6)+2csc(pi/3)

    Thanks a TONNE
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  2. #2
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    I have tryed to solve them the best I could...

    Im assuming that the first one is 0 because they are the inverse of eachother.

    The second one I got 2.73 by converting tan(pi/6) to sqr3/3 and csc to 2/sqr3, then solving accordingly.

    Does this sound correct?
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  3. #3
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    Quote Originally Posted by gabrie30 View Post

    How do you find the exact value of these expressions?

    sin^2 15(degrees) cot^2 75(degrees) + cos^2 15(degrees) tan^2 75(degrees)
    Note that:
     sin(15^{\circ}) = sin(45^{\circ}-30^{\circ})
    cot(75^{\circ}) = cot (45^{\circ} + 30^{\circ})<br />

    Use the sum-difference formulas to solve them. (i.e. sin(a-b) = sin(a)cos(b)-cos(a)sin(b))



    1-tan(pi/6)+2csc(pi/3)
     tan \frac{\pi}{6} has exact value  \frac{1}{\sqrt3}

     cosec \frac{\pi}{3} = \frac{1}{sin \frac{\pi}{3}}\Rightarrow sin\frac{\pi}{3} has exact value  \frac{\sqrt3}{2}
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  4. #4
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    Hello, gabrie30!

    Welcome aboard!


    Find the exact value: . \sin^2\!15^o\cot^2\!75^o + \cos^2\!15^o \tan^2\!75^o
    Use the Double Angle identities:

    . . \begin{array}{ccccccc}\sin^2\!\theta &=& \dfrac{1-\cos2\theta}{2} & & \cos^2\!\theta &=& \dfrac{1 + \cos2\theta}{2} \\ \\[-3mm] \tan^2\!\theta &=& \dfrac{1-\cos2\theta}{1 + \cos2\theta} & & \cot^2\theta &=& \dfrac{1 + \cos2\theta}{1-\cos2\theta} \end{array}


    We have: . \frac{1-\cos30^o}{2}\cdot\frac{1 + \cos150^o}{1-\cos150^o} + \frac{1+\cos30^o}{2}\cdot\frac{1 - \cos150^o}{1 + \cos150^o}


    Since \cos150^o \:=\:\text{-}\cos30^p, we have: . \frac{1-\cos30^o}{2}\cdot\frac{1-\cos30^o}{1+\cos30^o} + \frac{1+\cos30^o}{2}\cdot\frac{1+\cos30^o}{1-\cos30^o}

    . . which simplfies to: . \frac{(1-\cos30^o)^3 + (1+\cos30^o)^3}{2(1-\cos^230^o)} \;=\;\frac{1+3\cos^2\!30^o}{\sin^2\!30^o}


    Therefore, we have: . \frac{1 + 3\left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{1}  {2}\right)^2} \;=\;\frac{1 + 3\cdot\frac{3}{4}}{\frac{1}{4}} \;=\;\boxed{13}




    1-\tan\left(\frac{\pi}{6}\right) + 2\csc\left(\frac{\pi}{3}\right)
    These are standard angles . . .
    We are expected to know that: . \begin{array}{ccc}\tan\left(\dfrac{\pi}{6}\right) &=& \dfrac{1}{\sqrt{3}} \\ \\[-3mm] \csc\left(\dfrac{\pi}{3}\right) &=&\dfrac{2}{\sqrt{3}} \end{array}

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