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Math Help - Finding Exact value of expression Analytical Trig

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    Finding Exact value of expression Analytical Trig

    I am kinda lost on this guys. The question is:
    Find the exact value of the expression
    Cos(300+45)

    I am prty sure u have to use the angle formula, and i plugged it in, but i am lost now. anyone help me?
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  2. #2
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    Quote Originally Posted by robdarftw View Post
    I am kinda lost on this guys. The question is:
    Find the exact value of the expression
    Cos(300+45)

    I am prty sure u have to use the angle formula, and i plugged it in, but i am lost now. anyone help me?
    300 = (270+30)

    so we have cos((270+30)+45) = cos(270+30)cos(45) - sin(270+30)sin(45)

    cos(45) = sin(45) = \frac{1}{\sqrt2}

    \frac{cos(270+30)-sin(270+30)}{\sqrt2}

    using cos(A+B) and sin(A+B) with A=270 and B=30 gives:

    \frac{[cos(270)cos(30) - sin(270)sin(30)] - [sin(270)cos(30) + cos(270)sin(30]}{\sqrt2}

    cos(270) = 0

    sin(270) = -1

    cos (30) = \frac{1}{2}

    sin(30) = \frac{\sqrt3}{2}


    <br />
\frac{[0+\frac{\sqrt3}{2}] - [\frac{\sqrt3}{4}+ 0]}{\sqrt2}

    \frac{\sqrt3}{2} = \frac{2\sqrt3}{4}:

    \frac{2\sqrt3 - \sqrt3}{4\sqrt2} = \frac{\sqrt3}{4\sqrt2} = \frac{\sqrt6}{8}

    perhaps there is a shorter way?
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  3. #3
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    Quote Originally Posted by robdarftw View Post
    I am kinda lost on this guys. The question is:
    Find the exact value of the expression
    Cos(300+45)

    I am prty sure u have to use the angle formula, and i plugged it in, but i am lost now. anyone help me?
    OK, clearly this is \cos{345^\circ}.

    We could write this as \cos{(360^\circ - 15^\circ)}. So the angle is 15^\circ, just in the 4th quadrant. Remembering that cosine is positive in the fourth quadrant, we get

    \cos{(360^\circ - 15^\circ)} = \cos{15^\circ}

    This can be evaluated using a half-angle formula.

    Remember that \cos{\frac{\theta}{2}} = \sqrt{\frac{1 + \cos{\theta}}{2}}.

    So \cos{15^\circ} = \cos{\frac{30^\circ}{2}}

     = \sqrt{\frac{1 + \cos{30^\circ}}{2}}

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

     = \sqrt{\frac{1}{2} + \frac{\sqrt{3}}{4}}.


    So \cos{(300^\circ + 45^\circ)} = \sqrt{\frac{1}{2} + \frac{\sqrt{3}}{4}}.
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