Question

cos^3x + cos^3(2π/3+x) +cos^3(2π/3-x)=3/4cos3x

Answer 1

1)As 4π/3 = π + π/3 and 2π/3 = π - π/3,

as well cos(π + t) = cos(π - t) = -cos(t),

cos(x + 4π/3) = -cos(π/3 + x) and cos(x + 2π/3) = -cos(π/3 - x)

ii) So of the above, cos³(x + 2π/3) + cos³(x + 4π/3) = -{cos³(π/3 + x) + cos³(π/3 - x)}

Applying the identity, a³ + b³ = (a + b)³ - 3ab(a + b),

-{cos³(π/3 + x) + cos³(π/3 - x)} =

= -[{cos(π/3 + x) + cos(π/3 - x)}³ -3cos(π/3 + x)*cos(π/3 - x){cos(π/3 + x) + cos(π/3 - x)}]

= -[{2cos(π/3)*cos(x)}³ - 3{cos²x - sin²(π/3)}{2cos(π/3)*cos(x)}]

[Application: cos(A + B) + cos(A- B) = 2cos(A)*cos(B); cos(A + B)*cos(A - B) = cos²A - sin²B]

= -[cos³x - 3(cos²x - 3/4)*cos(x)] = -[cos³x - 3cos³x + (9/4)*cos(x)]

= 2cos³x - (9/4)*cos(x)

iii) So of the above left side = cos³x + 2cos³x - (9/4)*cos(x) = 3cos³x - (9/4)*cos(x)

= (3/4){4cos³x - 3*cos(x)} [Identity: 4cos³x - 3*cos(x) = cos(3x)]

= (3/4)*cos(3x) [Proved]

EDIT:

Just now I could solve the same in another method, which I feel simpler than this one.

i) As explained we have,

cos³x + cos³(x + 2π/3) + cos³(x + 4π/3) = cos³x - cos³(π/3 - x) - cos³(π/3 + x)

ii) cos(3A) = 4cos³A - 3cos(A)

==> cos³A = (1/4){cos(3A) + 3cos(A)}

Applying this,

cos³x = (1/4){cos(3x) + 3cos(x)}

-cos³(π/3 - x) = -(1/4){cos(π - 3x) + 3cos(π/3 - x)} = (1/4){cos(3x) - 3cos(π/3 - x)}

-cos³(π/3 + x) = -(1/4){cos(π + 3x) + 3cos(π/3 + x)} = (1/4){cos(3x) - 3cos(π/3 + x)}

Adding all the above,

Left side = (1/4)[3cos(3x) + 3cos(x) - 3{cos(π/3 + x) + cos(π/3 - x)}]

= (1/4)[3cos(3x) + 3cos(x) - 3{2cos((π/3)*cos(x)]

= (1/4)[3cos(3x) + 3cos(x) - 3cos(x)]

= (3/4)*cos(3x) [Proved]

Answer 2

Hey

i) As 4π/3 = π + π/3 and 2π/3 = π - π/3,

as well cos(π + t) = cos(π - t) = -cos(t),

cos(x + 4π/3) = -cos(π/3 + x) and cos(x + 2π/3) = -cos(π/3 - x)

ii) So of the above, cos³(x + 2π/3) + cos³(x + 4π/3) = -{cos³(π/3 + x) + cos³(π/3 - x)}

Applying the identity, a³ + b³ = (a + b)³ - 3ab(a + b),

-{cos³(π/3 + x) + cos³(π/3 - x)} =

= -[{cos(π/3 + x) + cos(π/3 - x)}³ -3cos(π/3 + x)*cos(π/3 - x){cos(π/3 + x) + cos(π/3 - x)}]

= -[{2cos(π/3)*cos(x)}³ - 3{cos²x - sin²(π/3)}{2cos(π/3)*cos(x)}]

[Application: cos(A + B) + cos(A- B) = 2cos(A)*cos(B); cos(A + B)*cos(A - B) = cos²A - sin²B]

= -[cos³x - 3(cos²x - 3/4)*cos(x)] = -[cos³x - 3cos³x + (9/4)*cos(x)]

= 2cos³x - (9/4)*cos(x)

iii) So of the above left side = cos³x + 2cos³x - (9/4)*cos(x) = 3cos³x - (9/4)*cos(x)

= (3/4){4cos³x - 3*cos(x)} [Identity: 4cos³x - 3*cos(x) = cos(3x)]

= (3/4)*cos(3x) [Proved]