identify the identities: sin3=(sinx)(3-4sin^2x) & solve algebraically in int. [0,2pi): cos2x+cos4x=0?

int. = interval

please show all work and how you got it.

thank you!

Update:

oops, on the first one, it is suppose to be sin3x=...

2 Answers

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  • 1 decade ago
    Favorite Answer

    Here's the identity one:

    LHS = sin 3x

    = sin(2x + x)

    = sin 2x cosx + cos 2x sinx

    = 2 sinx cosx cosx + (2cos² x - 1) sinx

    = 2 sinx cos² x + 2 sinx cos² x - sinx

    = 4 sinx cos² x - sinx

    = 4 sinx (1 - sin² x) - sinx

    = 4 sinx - 4 sin³ x - sin x

    = 3sinx - 4 sin³ x

    = sinx (3 - 4 sin² x)

    = RHE

    and the second one:

    cos 2x + cos 4x = 0

    2 cos² x - 1 + cos (2x + 2x) = 0

    2 cos² x - 1 + cos 2x cos 2x - sin 2x sin 2x = 0

    2 cos² x - 1 + (2 cos² x - 1)² - (2 sin x cos x)² = 0

    2 cos² x - 1 + 4 cos^4 x - 4 cos² x + 1 - 4 sin² x cos² x = 0

    4 cos^4 x - 2 cos² x - 4(1 - cos² x) cos² x = 0

    4 cos^4 x - 2 cos² x - 4 cos² x + 4 cos^4 x = 0

    8 cos^4 x - 6 cos² x = 0

    2cos² x (4 cos² x - 3) = 0

    So 2 cos² x = 0 OR 4 cos² x - 3 = 0

    cos² x = 0 OR cos² x = 3/4

    cos x = 0 OR cos x = ± √3/2

    If cos x = 0, x = π/2 or x = 3π/2

    If cos x = √3/2, x = π/6 or x = 2π - π/6 = 11π/6 [Because cos is positive in the 1st and 4th quadrants]

    If cos x = -√3/2, x = π - π/6 = 5π/6 or x = π + π/6 = 7π/6 [because cos is negative in the 2nd and 3rd quadrants]

    so for the given domain x = π/6 or π/2 or 5π/6 or 7π/6 or 3π/2 or 11π/6

  • joles
    Lv 4
    4 years ago

    confident. then denote cos x by y and get the equation 2y^2-a million = y or 2y^2 - x - a million = 0. then resolve it by quadratic formula. you're able to be able to desire to nonetheless ensure if the result's between -a million and a million ( as cos x must be). If the result isn't between - a million and a million the eqution does not have recommendations. ultimately you will discover the recommendations of the type x = 2*pi*n + arcos(y), the place n is the set of integers

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