Note: z = cis(@) here.
I worked out that
z + z^-1 = 2cos(@) ---1
(z + z^-1)^3 = z^3 + z^-3 + 3(z+z^-1) --- 2
Now I need to prove cos(3@) = 4cos^3(@) - 3cos(@)
I try to sub in 2cos(@) into 2 for (z + z^-1), and fiddle around, but am unable to prove the above.
Any suggestions? Thanks.
Hello, classicstrings!
I could help more if I knew what the question is . . .
What are we trying to prove?
Cube both sides: .[z + z^{-1}]³ .= .8·cos³θNote: z = cisθ here.
I worked out that: .z + z^{-1} .= /2·cosθ . . . . good!
And we have: .z^3 + z^{-3} + 3[z + z^{-1}] .= .8·cos³θ
. . which becomes: .z^3 + z^{-3} + 3·2·cosθ .= .8·cos³θ
. . and we have: .z^3 + z^{-3} .= .8·cos³θ - 6·cosθ
This is a standard multiple-angle identity.Now I need to prove: .cos(3θ) .= .4·cos³θ - 3·cosθ . . why?
To prove it, we need four more identities:
. . cos(A + B) .= .cos(A)·cos(B) - sin(A)·sin(B)
. . . . .sin(2A) .= .2·sin(A)·cos(A)
- - . . cos(2A) .= .2·cos²(A) - 1
. . . . .sin²(A) .= .1 - cos²(A)
We have: .cos(3θ) .= .cos(2θ + θ)
. . . . . . . . . . . . . .= .cos(2θ)·cos(θ) - sin(2θ)·sin(θ)
. . . . . . . . . . . . . .= .[2·cos²(θ) - 1]·cos(θ) - [2·sin(θ)·cos(θ)]·sin(θ)
. . . . . . . . . . . . . .= .2·cos³(θ) - cos(θ) - 2·cos(θ)·sin²(θ)
. . . . . . . . . . . . . .= .2·cos³(θ) - cos(θ) - 2·cos(θ)[1 - cos²(θ)]
. . . . . . . . . . . . . .= .2·cos³(θ) - cos(θ) - 2·cos(θ) + 2·cos³(θ)
. . . . . . . . . . . . . .= .4·cos³(θ) - 3·cos(θ)
. . . . . ta-DAA!
Hello Soroban!
The point of the question was to use the previous two results to prove the cos(3θ) .= .4·cos³θ - 3·cosθ .
Looking over CB's solution I was able to get it.
Now I have to consider z - z^-1, to find an expression for sin3@ in terms of sin@.
So z - z^-1 = 2isin@
8isin^3@ = z^3-z^-3 - 3(z - z^-1)
Hello again, classicstrings!
Then you were only a few steps away . . . and so was I.
Note that:We have: .z + z^{-1} .= .2·cosθ
Cube both sides: .[z + z^{-1}]³ .= .8·cos³θ
And we have: .z^3 + z^{-3} + 3[z + z^{-1}] .= .8·cos³θ
. . which becomes: .z^3 + z^{-3} + 3·2·cosθ .= .8·cos³θ
. . and we have: .z^3 + z^{-3} .= .8·cos³θ - 6·cosθ .[1]
. . z^3 . = .cos3θ + i·sin3θ
. . z^-3 .= .cos3θ - i·sin3θ
Add: . z^3 + z^{-3} .= .2·cos3θ .[2]
Equate [1] and [2]: .2·cos3θ .= .8·cos³θ - 6·cosθ . → . cos3θ .= .4·cos³θ - 3·cosθ