Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Is cos^2 (2x)-sin^2(2x)=0 an identity?

At first glance, I don’t think it is, but I’m not sure. How would I approach this question? 

Thank you for your time. 

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  • 1 month ago

    There is difference in an equation and an identity. 

    In an Equation all the variables can have a limited number of values. where as in identity. the relation is satisfied by any value given to the variable(s). 

    For example consider your problem -

    cos² 2x - sin² 2x = 0

    It has the highest power = 2, Now if the given relation is satisfied by assigning more than two ( say 3) values of x, it is an identity. 

    If this relation is satisfied with three values of x, then it is an Identity. Lets see -

    Put x = 30°, 45°, 60°, 90°

    cos²(90°) - sin²(90°) = 0 - 1  =  - 1 

    It means that x = 45° DOES NOT SATISFY the relation. 

    Similarly =>  cos²(30°) - sin²(30°)  ≠  0

    cos²(45°) - sin²(45°) = 0  .................... Satisfies.

    cos²(60°) - sin²(60°) ≠ 0

    Hence --

    cos² 2x - sin² 2x = 0  is an equation and not an identity.

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  • 1 month ago

    Remember the Trig. Identity

    Cos(2z) = Cos^(z0 - Sin^2(Z)

    Hence it follows

    Cos(4x) = Cos(2x + 2x)

    Cos(2x + 2x) = Cos^(2x) - Sin^2(2x)

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  • 1 month ago

    cos²(2x) - sin²(2x) = 0 ← not an identity but an equation

    cos²(2x) = sin²(2x)

    sin²(2x)/cos²(2x) = 1

    tan²(2x) = 1

    tan(2x) = ± 1

    First case: tan(2x) = 1

    2x = (π/4) + kπ → where k is an integer

    x = (π/8) + k.(π/2)

    Second case: tan(2x) = - 1

    2x = [π - (π/4)] + kπ → where k is an integer

    2x = (3π/4) + kπ

    x = (3π/8) + k.(π/2)

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  • sepia
    Lv 7
    1 month ago

    cos^2 (2x) - sin^2 (2x) = 0

    cos(4 x) = 0

    Solution:

    x = (π n)/4 - π/8, n element Z

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  • Pope
    Lv 7
    1 month ago

    No, it is not an identity. For the simplest of counterexamples, let x = 0.

    x = 0

    cos²(2x) - sin²(2x)

    = cos²(0) - sin²(0)

    = 1 + 0

    = 1

    ≠ 0

    The equation does have solutions though.

    Let cos²(2x) - sin²(2x) = 0.

    sin²(2x) = cos²(2x)

    sin²(2x) / cos²(2x) = 1

    tan²(2x) = 1

    tan(2x) = ±1

    2x = (1 + 2k)π/4

    x = (1 + 2k)π/8, for any integer k

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  • TomV
    Lv 7
    1 month ago

    Let x = 0 which results in:

    cos²(0) - sin²(0) = 0

    1 - 0 = 0

    1 = 0

    This is a false statement. If the original equation is an identity, it must be true for all values of x. It is not. Therefore it is not an identity.

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  • 1 month ago

    cos^2(2x) - sin^2(2x) = 0

    cos^2(2x) = sin^2(2x)

    cos(2x) = +/- sin(2x)

    sin(2x)/cos(2x) = +/- 1

    tan(2x) = +/- 1

    2x = (pi*n/2) - (pi/4) radians, for any integer n

    x = (pi*n/4) - (pi/8) radians, for any integer n

    x = 22.5n degrees, for any odd number n

    Any other value for x doesn't work, so therefore that's NOT an identity.

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  • It's not an identity because this isn't true for all values of x.  However, we can solve for x

    cos(2x)^2 - sin(2x)^2 = 0

    cos(2 * 2x) = 0

    cos(4x) = 0

    4x = pi/2 + pi * k

    4x = (pi/2) * (1 + 2k)

    x = (pi/8) * (1 + 2k)

    x = pi/8 , 3pi/8 , 5pi/8 , 7pi/8 , 9pi/8 , 11pi/8 , 13pi/8 , 15pi/8

    k is an integer

    cos(2x)^2 - sin(2x)2 = 0

    cos(2x)^2 = sin(2x)^2

    1 = tan(2x)^2

    -1 , 1 = tan(2x)

    pi/4 + pi * k , 3pi/4 + pi * k = 2x

    pi/4 + (pi/2) * k = 2x

    (pi/4) * (1 + 2k) = 2x

    (pi/8) * (1 + 2k) = x

    Which is exactly what we had before.

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