Double Angle Identities

majtata

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May 19, 2007
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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.
 
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
 
Hello, majtata!

Solve: 3sin6xcsc2x=4\displaystyle \,3\cdot\sin6x\cdot\csc2x\:=\:4

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

You know the double-angle identity: sin2θ=2sinθcosθ\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.

sin3θ=sin(2θ+θ)\displaystyle \sin3\theta\:=\:\sin(2\theta\,+\,\theta)
. . . . =sin2θcosθ+sinθcos2θ\displaystyle = \:\sin2\theta\cdot\cos\theta\,+\,\sin\theta\cdot\cos2\theta
. . . . \(\displaystyle = \:(2\cdot\sin\theta\cdot\cos\theta)\cdot\cos\theta\,+\,\sin\theta(1\,-\,2\cdot\sin^2\theta)\)
. . . . =2sinθcos2θ+sinθ2sin3θ\displaystyle =\:2\cdot\sin\theta\cdot\cos^2\theta\,+\,\sin\theta\,-\,2\cdot\sin^3\theta
. . . . =2sinθ(1sin2θ)+sinθ2sin3θ\displaystyle =\:2\cdot\sin\theta(1\,-\,\sin^2\theta)\,+\,\sin\theta\,-\,2\cdot\sin^3\theta
. . . . =2sinθ2sin3θ+sinθ2sin3θ\displaystyle =\:2\cdot\sin\theta\,-\,2\cdot\sin^3\theta\,+\,\sin\theta\,-\,2\cdot\sin^3\theta
. . . . =3sinθ4sin3θ\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: sin6x=3sin2x4sin32x\displaystyle \:\sin6x \:=\:3\cdot\sin2x\,-\,4\cdot\sin^32x

The equation becomes: 3(3sin2x4sin32x)1sin2x=4\displaystyle \:3\left(3\cdot\sin2x\,-\,4\cdot\sin^32x\right)\,\cdot\,\frac{1}{\sin2x}\:=\:4

. . which reduces: 3(34sin22x)=4  \displaystyle \:3(3\,-\,4\cdot\sin^22x)\:=\:4\; **

. . and simplifies to: sin32x=512\displaystyle \:\sin^32x\:=\:\frac{5}{12}

And we have: sin2x=5123\displaystyle \:\sin2x\:=\:\sqrt[3]{\frac{5}{12}}

. . . . . . . . . . . . .2x=arcsin(5123)\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 sin2x\displaystyle \sin2x.

In the original equation, we see that: sin6x=0\displaystyle \sin6x\,=\,0 is not a solution.
. . Hence, x\displaystyle x is not equal to: 0,π6,π3,π2,2π3,5π6,π,\displaystyle \,0,\:\frac{\pi}{6},\:\frac{\pi}{3},\:\frac{\pi}{2},\:\frac{2\pi}{3},\:\frac{5\pi}{6},\:\pi,\:\cdots
Therefore: sin2x0\displaystyle \sin2x\,\neq\,0

 
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