Double Angle Identities

[majtata]

I am currently doing a past paper which ends with a trigonometric identities question which requires solving:

3sin6X.cosec2X = 4

cosec2X is simply 1/sin2X but what is 3sin6X?

I know sin2X = 2sinXcosX but don't understand how to use this identity in this context?

Any help would be appreciated.

 

[morson]

To make things look a bit neater, let u = 2x:

3sin(3u) * cosec(u) = 4

Now, let's take a look at de Moivre's theorem, which states:

[cos(u) + i*sin(u)]^3 = cos(3u) + i*sin(3u)

Expanding, [cos(u) + i*sin(u)]^3 = (cosu)^3 + 3*(cosu)^2*i*sinu - 3(cosu)(sinu)^2 - i(sinu)^3
= [(cosu)^3 - 3(cosu)(sinu)^2] + i[3(sinu)(cosu)^2 - (sinu)^3]

Comparing real and imaginary coefficients, we come up with:

sin(3u) = 3(sinu)(cosu)^2 - (sinu)^3

So substituting this back into our equation:

[9(sinu)(cosu)^2 - 3(sinu)^3] * cosec(u) = 4

Hence, 9(cosu)^2 - 3(sinu)^2 = 4, as cosec(u) = 1/sin(u)

Therefore: 12(cosu)^2 = 7

 

[soroban]

Hello, majtata!

Solve: \(\displaystyle \,3\cdot\sin6x\cdot\csc2x\:=\:4\)

\(\displaystyle \csc2x\:=\:\frac{1}{sin2x}\), but what is \(\displaystyle \sin6x\) ?

You know the double-angle identity: \(\displaystyle \,\sin2\theta\:=\:2\cdot\sin\theta\cdot\cos\theta\) . . . Good!

We need a "triple-angle identity".
. . If you don't know one, we can derive it.

\(\displaystyle \sin3\theta\:=\:\sin(2\theta\,+\,\theta)\)
. . . . \(\displaystyle = \:\sin2\theta\cdot\cos\theta\,+\,\sin\theta\cdot\cos2\theta\)
. . . . \(\displaystyle = \2\cdot\sin\theta\cdot\cos\theta)\cdot\cos\theta\,+\,\sin\theta(1\,-\,2\cdot\sin^2\theta)\)
. . . . \(\displaystyle =\:2\cdot\sin\theta\cdot\cos^2\theta\,+\,\sin\theta\,-\,2\cdot\sin^3\theta\)
. . . . \(\displaystyle =\:2\cdot\sin\theta(1\,-\,\sin^2\theta)\,+\,\sin\theta\,-\,2\cdot\sin^3\theta\)
. . . . \(\displaystyle =\:2\cdot\sin\theta\,-\,2\cdot\sin^3\theta\,+\,\sin\theta\,-\,2\cdot\sin^3\theta\)
. . . . \(\displaystyle =\:3\cdot\sin\theta\,-\,4\cdot\sin^3\theta\)

Hence: \(\displaystyle \:\fbox{\sin3\theta\:=\:3\cdot\sin\theta\,-\,4\cdot\sin^3\theta}\)



Back to the problem: \(\displaystyle \:\sin6x \:=\:3\cdot\sin2x\,-\,4\cdot\sin^32x\)

The equation becomes: \(\displaystyle \:3\left(3\cdot\sin2x\,-\,4\cdot\sin^32x\right)\,\cdot\,\frac{1}{\sin2x}\:=\:4\)

. . which reduces: \(\displaystyle \:3(3\,-\,4\cdot\sin^22x)\:=\:4\;\) **

. . and simplifies to: \(\displaystyle \:\sin^32x\:=\:\frac{5}{12}\)

And we have: \(\displaystyle \:\sin2x\:=\:\sqrt[3]{\frac{5}{12}}\)

. . . . . . . . . . . . .\(\displaystyle 2x\:=\:\arcsin\left(\sqrt[3]{\frac{5}{12}}\right)\)

Therefore: . . . . \(\displaystyle \L x\:=\:\frac{1}{2}\cdot\arcsin\left(\sqrt[3]{\frac{5}{12}}\right)\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

We can cancel \(\displaystyle \sin2x\).

In the original equation, we see that: \(\displaystyle \sin6x\,=\,0\) is not a solution.
. . Hence, \(\displaystyle x\) is not equal to: \(\displaystyle \,0,\:\frac{\pi}{6},\:\frac{\pi}{3},\:\frac{\pi}{2},\:\frac{2\pi}{3},\:\frac{5\pi}{6},\:\pi,\:\cdots\)
Therefore: \(\displaystyle \sin2x\,\neq\,0\)