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Trigonoemtric Identity

  1. Dec 2, 2014 #1
    Why ##\sin^2 2x + \cos^2 2x = 1##?

    Will

    ##\sin^2 3x + \cos^2 3x##

    or

    ##\sin^2 4x + \cos^2 4x##

    and so on, be = 1?

    How to proof this?
     
  2. jcsd
  3. Dec 2, 2014 #2

    Bystander

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    You know the definitions of the functions sin(x) and cos(x); you also are familiar with the Pythagorean theorem.
     
  4. Dec 2, 2014 #3
    I am familiar with the Pythagorean theorem but I don't know the definitions of the functions sin(x) and cos(x).

    Explain please.
     
  5. Dec 2, 2014 #4

    Bystander

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    You have got to be kidding me. Given a right triangle, we'll plot its vertices on Cartesian coordinates (0,0), (1,0), and (1,y) where y = 0 to +∞, and the interior angle at (0,0), α = 0 to 90°, or 0 to π/2 radians, sin(α) ≡ y/(x2 + y2)1/2, or the ratio of the length of the side of the triangle opposite ∠α to the length of the hypotenuse of the right triangle. cos(α) ≡ the ratio of the length of the base of the triangle, x, to the length of the hypotenuse.

    Now, give it a try.

    Edited to include sqrt --- my bad.
     
    Last edited: Dec 2, 2014
  6. Dec 2, 2014 #5
    Why ##\sin (a) = \frac{y}{x^2 + y^2}##?
     
  7. Dec 2, 2014 #6

    Mark44

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    The identity is ##sin^2(\text{whatever}) + cos^2(\text{whatever}) = 1##. Do you see how this fits with the questions you asked in post #1?
    It's not - there's a radical that's missing. It should be
    ##\sin (a) = \frac{y}{\sqrt{x^2 + y^2}}##

    If you don't understand where that comes from, you need to review right triangle trigonometry.
     
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