Thread: My Problem

Guys can anyone help me with this problem:
If Sin(Heata)=3 X Sin (Heata + 2 X Alpha)
Prove That: Tan(Heata + Alpha) + 2 X Tan (Alpha) =0

Hello, omega2692!

You have a new spelling for "theta"!

There must be a simpler solution; I hope someone can find one.

If sin(θ) = 3·sin(θ+ 2α)

prove that: .tan(θ + α) + 2·tan(α) .= .0 .[A]

We are given: .sin(θ) . = . 3·sin(θ + 2α) . = . 3·sin(θ)·cos(2α) + 3·cos(θ)·sin(α)

Then: . sin(θ) - 3·sin(θ)·cos(2α) . = . 3·cos(θ)·sin(2α)

Factor: . sin(θ) [1 - 3·cos(2α)] . = . 3·cos(θ)·sin(2α)

. . . . . . . . . . . sin(θ) . . . . . .3·sin(2α)
and we have: . -------- . = . --------------- . = . tan(θ)
. . . . . . . . . . . cos(θ) . . . .1 - 3·cos(2α)


. . . . . . . . . . . . . . 3[2·sin(α)·cos(α)] . . . . 3·sin(α).cos(α)
Then: . tan(θ) . = . ---------------------- . = . ----------------- .[1]
. . . . . . . . . . . . . .1 - 3[2cos²(α) - 1] . . . . . 2 - 3cos²(α)



. . . . . . . . . . . . . . . . . . . . . . . tan(θ) + tan(α)
We know that: . tan(θ + α) .= . -------------------- .[2]
. . . . . . . . . . . . . . . . . . . . . . .1 - tan(θ)·tan(α)


Substitute [1] into [2]:

. . . . . . . . . . . . . . .3·sin(α)·cos(α) . sin(α)
. . . . . . . . . . . . . . ----------------- + --------
. . . . . . . . . . . . . . . 2 - 3·cos²α . . . cos(α)
. . tan(θ + α) . = . ---------------------------------
. . . . . . . . . . . . . . . . .3·sin(α)·cos(α) .sin(α)
. . . . . . . . . . . . . .1 + ------------------.-------
. . . . . . . . . . . . . . . . . .2 - 3·cos(α) . .cos(α)


. . . . . . . . . . . . . . . 3·sin(α).cos²(α) + 2·sin(α) - 3·sin(α)·cos²(α)
which simplifies to: . ----------------------------------------------------
. . . . . . . . . . . . . . . . . 2·cos(α) - 3·cos³(α) - 3·sin²(α)·cos(α)


. . . . . . . . . . . . . . . . . . . 2·sin(α)
and further simplifies to: . ----------- . = . -2·tan(α)
. . . . . . . . . . . . . . . . . . . - cos(α)



Substitute into [A]: . -2·tan(α) + 2·tan(α) . = . 0