Trigonometric Values As a Function of x...1

Discussion in 'Other Pre-University Math' started by nycmathguy, Feb 4, 2022.

  1. nycmathguy

    nycmathguy

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    David Cohen Textbook Questions.
    Enjoy.

    IMG_20220204_134648.jpg
     
    nycmathguy, Feb 4, 2022
    #1
  2. nycmathguy

    MathLover1

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    sin^2(theta)=(2x)^2
    sin^2(theta)=4x^2

    cos^2(theta)=3^2
    cos^2(theta)=9

    tan^2(theta)=sin^2(theta)/cos^2(theta)=4x^2/9
     
    MathLover1, Feb 4, 2022
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Not as hard as it looks.
     
    nycmathguy, Feb 5, 2022
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  4. nycmathguy

    Country Boy

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    The legs have lengths 2x and 3 so, by the Pythagorean formula, the hypotenuse has length sqrt(4x^2+ 9).

    sin(theta)= 2x/sqrt(4x^2+ 9) so sin^2(theta)= 4x^2/(4x^2+ 9)

    cos(theta)= 3/sqrt(4x^2+ 9) so cos^2(theta)= 9/(4x^2+ 9)

    tan(theta)= 2x/3 so tan^2(theta)= 4x^2/9.
     
    Country Boy, Feb 5, 2022
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  5. nycmathguy

    nycmathguy

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    Cool.
     
    nycmathguy, Feb 5, 2022
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