Trigonometric Prove...2

[nycmathguy]

David Cohen

 

[MathLover1]

suppose that

x+1/x=2cos(θ)

show that x^3+1/x^3=2cos(3θ)

Given,
x+1/x=2cos(θ)

=>(x+1/x)^3=(2cos(θ) )^3

(x+1/x)^3=(2cos(θ) )^3

x^3 + 1/x^3 + 3 x + 3/x=8cos^3(θ)

x^3 + 1/x^3 + 3(x + 1/x)=8cos^3(θ) ......substitute given value for (x+1/x)
x^3 + 1/x^3 + 3*2cos(θ)=8cos^3(θ)

x^3 + 1/x^3 + 6cos(θ)=8cos^3(θ)

x^3 + 1/x^3 =8cos^3(θ)-6cos(θ)

x^3 + 1/x^3 =2cos(theta)(4cos^2(θ)-3)

x^3 + 1/x^3 =2cos(θ)(2cos(2 θ) - 1)

x^3 + 1/x^3 =2cos3θ.

 

[nycmathguy]

suppose that

x+1/x=2cos(θ)

show that x^3+1/x^3=2cos(3θ)

Given,
x+1/x=2cos(θ)

=>(x+1/x)^3=(2cos(θ) )^3

(x+1/x)^3=(2cos(θ) )^3

x^3 + 1/x^3 + 3 x + 3/x=8cos^3(θ)

x^3 + 1/x^3 + 3(x + 1/x)=8cos^3(θ) ......substitute given value for (x+1/x)
x^3 + 1/x^3 + 3*2cos(θ)=8cos^3(θ)

x^3 + 1/x^3 + 6cos(θ)=8cos^3(θ)

x^3 + 1/x^3 =8cos^3(θ)-6cos(θ)

x^3 + 1/x^3 =2cos(theta)(4cos^2(θ)-3)

x^3 + 1/x^3 =2cos(θ)(2cos(2 θ) - 1)

x^3 + 1/x^3 =2cos3θ.

I am convinced that there are no math problems you cannot solve.