Anonymous
Anonymous asked in Education & ReferenceHomework Help · 10 years ago

Double Angle Identities Calculus Help?

My teacher wasn't here today and left us new work. I did most of it expect three

Prove the identity- cos6x = 2cos^2 3x-1 (^2 means raised to the 2nd power)

Solve algebraically for exact solutions- cos2x + sinx =0 & cosx + cos3x = 0

Please show all work thanks you!!

1 Answer

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  • Anonymous
    10 years ago
    Favorite Answer

    1. LHS . . . . RHS

    cos(6x) = 2cos²(3x) - 1

    * * *

    Rewrite cos(6x) as:

    cos(2(3x))

    Recall that:

    cos(2x) = 2cos²(x) - 1

    = 1 - 2sin²(x)

    = cos²(x) - sin²(x)

    Finally, by double-angle identity (letting x = 3x), we obtain:

    2cos²(3x) - 1

    ==> RHS

    2. cos(2x) + sin(x) = 0

    By double-angle identity, we obtain:

    1 - 2sin²(x) + sin(x) = 0

    -2sin²(x) + sin(x) + 1 = 0

    Then,

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

    2sin(x) + 1 = 0 and -sin(x) + 1 = 0

    sin(x) = -½ and sin(x) = 1

    x = arcsin(-½) and x = arcsin(1)

    x = -π/6, π/2

    Since the period is 2π, we obtain:

    x = -π/6 + 2n, π/2 + 2πn, where n is the integer

    3. cos(x) + cos(3x) = 0

    Recall that:

    cos(3x) = 4cos³(x) - 3cos(x)

    By triple-angle identity, we obtain:

    cos(x) + 4cos³(x) - 3cos(x) = 0

    4cos³(x) - 2cos(x) = 0

    Then,

    2cos(x)(2cos²(x) - 1) = 0

    2cos(x) = 0 and 2cos²(x) - 1 = 0

    cos(x) = 0 and cos(x) = ±√(½)

    cos(x) = 0 and cos(x) = √(½) and cos(x) = -√(½)

    x = arccos(0) and x = arccos(√(½)) and x = arccos(-√(½))

    Therefore,

    x = πn - π/2 where n is the integer

    I hope this helps!

    Source(s): Knowledge
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