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anonymous

  • one year ago

Find the sine, cosine, and tangent of 45 degrees. Sin 45 degrees = negative square root of 2 divided by 2, cos 45 degrees = negative square root of 2 divided by 2, tan 45 degrees = negative square root of 2 Sin 45 degrees = square root of 2 divided by 2, cos 45 degrees = square root of 2 divided by 2, tan 45 degrees = square root of 2 Sin 45 degrees = square root of 2 divided by 2, cos 45 degrees = square root of 2 divided by 2, tan 45 degrees = 1 Sin 45 degrees = square root of 2 divided by 2, cos 45 degrees = square root of 2 divided by 2, tan 45 degrees = −1

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  1. johnweldon1993
    • one year ago
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    Remember your special triangles, it can make a problem like this very quick A 45-45-90 triangle |dw:1444252285965:dw|

  2. johnweldon1993
    • one year ago
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    \[\large sin(\theta) = \frac{opposite}{hypotenuse}\] \[\large cos(\theta) = \frac{adjacent}{hypotenuse}\] \[\large tan(\theta) = \frac{opposite}{adjacent}\]

  3. anonymous
    • one year ago
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    oh now i remember!!! thank you so mcuh!

  4. johnweldon1993
    • one year ago
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    No problem :)

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