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Math Help - Trig question I need to understand this tonight!

  1. #1
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    Trig question I need to understand this tonight!

    Okay...

    Given that:
    sin cos
    66 degrees 0.9135 0.4067

    104 degrees 0.9703 -0.2419

    213 degrees -0.5446 -0.8387

    Find sin 81 degrees

    Hint: Double, Difference


    Apparently I don't understand the concept. Thanks!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by aigiqinf View Post
    Okay...

    Given that:
    sin cos
    66 degrees 0.9135 0.4067

    104 degrees 0.9703 -0.2419

    213 degrees -0.5446 -0.8387

    Find sin 81 degrees

    Hint: Double, Difference


    Apparently I don't understand the concept. Thanks!
    Find a way to get 81 to be a combination of 213,104,66 for example this wouldnt work but lets say you wanted 128 degrees
    128=213-\frac{66}{2}-\frac{104}{2} and once you find a way of doing that to get 81, use your trig identities to get them into a form you can use the given information with
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Find a way to get 81 to be a combination of 213,104,66 for example this wouldnt work but lets say you wanted 128 degrees
    128=213-\frac{66}{2}-\frac{104}{2} and once you find a way of doing that to get 81, use your trig identities to get them into a form you can use the given information with
    I've been trying to do that... I'll continue to do so. I'm pretty sure that that's not the way I'm supposed to do it.
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  4. #4
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    Quote Originally Posted by aigiqinf View Post
    Okay...

    Given that:
    sin cos
    66 degrees 0.9135 0.4067

    104 degrees 0.9703 -0.2419

    213 degrees -0.5446 -0.8387

    Find sin 81 degrees

    Hint: Double, Difference


    Apparently I don't understand the concept. Thanks!
    Sin(81) = Sin(213-66-66) = Sin((213-66)-66)

    Sin(213-66) = Sin213Cos66 - Cos213Cos66

    Cos(213-66) = Cos213Cos66 + Sin213Sin66

    Sin((213-66) - 66) = (Sin213Cos66 - Cos213Cos66)(Cos66) - (Cos213Cos66 + Sin213Sin66)(Sin66)
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  5. #5
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    Hello, aigiqinf!

    The hint refers to:

    . . a Difference identity: . \sin(A-B) \:=\:\sin(A)\cos(B) - \cos(A)\sin(B)
    . . two Double-Angle identities: . \begin{array}{ccc}\sin(2A) &=& 2\sin(A)\cos(A) \\ \cos(2A) &=&2\cos^2(A) - 1 \end{array}


    Given that: . \begin{array}{c|cc} & \text{sin} & \text{cos} \\ \hline<br />
66^o & 0.9135 &  0.4067 \\<br />
104^o & 0.9703 & \text{-}0.2419 \\<br />
213^o & \text{-}0.5446 & \text{-}0.8387  \end{array}

    Find \sin81^o

    Hint: Double, Difference
    We note that: . 81 \:=\:213 - 132

    So: . \sin81^o \;=\;\sin(213^o - 132^o) \;=\;\sin(213^o)\cos(132^o) - \cos(213^o)\sin(132^o)

    . . Since 132 \:=\:2(66), we have: . \sin(213^o)\overbrace{\cos(2\!\cdot\!66^o)} - \cos(213^o)\overbrace{\sin(2\!\cdot\!66^o)}

    . . . . . . . . . . . . . . . . = \;\sin(213^o)\overbrace{\bigg[2\cos^2(66^o) - 1\bigg]} - \cos(213^o)\overbrace{\bigg[2\!\cdot\!\sin(66^o)\!\cdot\!\cos(66^o)\bigg]}

    Now substitute the values given in the table . . .

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, aigiqinf!

    The hint refers to:

    . . a Difference identity: . \sin(A-B) \:=\:\sin(A)\cos(B) - \cos(A)\sin(B)
    . . two Double-Angle identities: . \begin{array}{ccc}\sin(2A) &=& 2\sin(A)\cos(A) \\ \cos(2A) &=&2\cos^2(A) - 1 \end{array}


    We note that: . 81 \:=\:213 - 132

    So: . \sin81^o \;=\;\sin(213^o - 132^o) \;=\;\sin(213^o)\cos(132^o) - \cos(213^o)\sin(132^o)

    . . Since 132 \:=\:2(66), we have: . \sin(213^o)\overbrace{\cos(2\!\cdot\!66^o)} - \cos(213^o)\overbrace{\sin(2\!\cdot\!66^o)}

    . . . . . . . . . . . . . . . . = \;\sin(213^o)\overbrace{\bigg[2\cos^2(66^o) - 1\bigg]} - \cos(213^o)\overbrace{\bigg[2\!\cdot\!\sin(66^o)\!\cdot\!\cos(66^o)\bigg]}

    Now substitute the values given in the table . . .

    Thank you!

    -.5446*(2(.4067^2-1)--.8387(2*.9135*.4067)=.9876 degrees
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