Processing math: 28%

Tuesday, June 16, 2015

Hardest Trigonometric Equation (First Solution)

Solve the trigonometric equation.

2cos(x5π12)6sin(x5π12)=2(sin(x52π3)sin(3x5+π6))

My solution:

By letting A=x5π12 use the sum-to-product formula to simplify the LHS of the equation, I get:

2cosA6sinA=2(sin(A7π12)sin(3A+5π12))=2(2cos(2Aπ12)sin(Aπ2))=4(cos(2Aπ12)sin(A+π2))=4cos(2Aπ12)cosA


4cos(2Aπ12)cosA=6sinA2cosA

Now, divide the left and right side of the equation by cosA and use the formulas of:

1.

cos2A=1tan2A1+tan2A,

2.

sin2A=2tanA1+tan2A, and

3.

sin(π12)=24(31);

cos(π12)=24(3+1)


to further simplify the equation, we get:

\displaystyle \dfrac{4\cos \left(2A-\frac{\pi}{12}\right)\cos A}{\cos A}=\dfrac{\sqrt{6}\sin A- \sqrt{2}\cos A}{\cos A}

\displaystyle 4\cos \left(2A-\frac{\pi}{12}\right)=\sqrt{6}\tan A- \sqrt{2}

\displaystyle 4\cos 2A \cos \left(\frac{\pi}{12}\right)+4\sin 2A \sin \left(\frac{\pi}{12}\right)=\sqrt{6}\tan A- \sqrt{2}

\displaystyle \small 4\left(\frac{1-\tan^2 A}{1+\tan^2 A}\right) \left(\dfrac{\sqrt{2}}{4}(\sqrt{3}+1)\right)+4\left(\frac{2\tan A}{1+\tan^2 A}\right) \left(\dfrac{\sqrt{2}}{4}(\sqrt{3}-1)\right)=\sqrt{6}\tan A- \sqrt{2}

\displaystyle \scriptsize \cancel{4}\left(\frac{1-\tan^2 A}{1}\right) \left(\dfrac{\sqrt{2}}{\cancel{4}}(\sqrt{3}+1)\right)+\cancel{4}\left(\frac{2\tan A}{1}\right) \left(\dfrac{\sqrt{2}}{\cancel{4}}(\sqrt{3}-1)\right)=(\sqrt{6}\tan A- \sqrt{2})(1+\tan^2 A)

\displaystyle \small \left(1-\tan^2 A\right) \left(\sqrt{2}(\sqrt{3}+1)\right)+\left(2\tan A\right) \left(\sqrt{2}(\sqrt{3}-1)\right)=\sqrt{2}(\sqrt{3}\tan A- 1)(1+\tan^2 A)

\displaystyle \small \left(1-\tan^2 A\right) \left(\cancel{\sqrt{2}}(\sqrt{3}+1)\right)+\left(2\tan A\right) \left(\cancel{\sqrt{2}}(\sqrt{3}-1)\right)=\cancel{\sqrt{2}}(\sqrt{3}\tan A- 1)(1+\tan^2 A)

\displaystyle \left(1-\tan^2 A\right) (\sqrt{3}+1)+\left(2\tan A\right)(\sqrt{3}-1)=(\sqrt{3}\tan A- 1)(1+\tan^2 A)

\displaystyle \sqrt{3}\tan^3 A+\sqrt{3}\tan^2 A+(2-\sqrt{3})\tan A-(2+\sqrt{3})=0 (*)

It's quite obvious that \displaystyle \tan A=1 is one of the solution to (*) since:

\displaystyle \begin{align*}\sqrt{3}+\sqrt{3}+(2-\sqrt{3})-(2+\sqrt{3})&=2\sqrt{3}+2-\sqrt{3}-2-\sqrt{3}\\&=0\end{align*}

We then use the long division to find the other two roots.

\displaystyle (\tan A-1)(\sqrt{3}\tan^2 A+2\sqrt{3}\tan A+2+\sqrt{3})=0

Since the discriminant of the quadratic expression that we found above \sqrt{3}\tan^2 A+2\sqrt{3}\tan A+2+\sqrt{3}:

\displaystyle \begin{align*}\text{discriminant}&=(-2\sqrt{3})^2-4(\sqrt{3})(2+\sqrt{3})\\&=12-12-8\sqrt{3}\\&=-8\sqrt{3}\\&\lt 0\end{align*}

is a negative value:, we can say the other two roots are imaginary.

Hence the solutions that we have found are

\displaystyle \tan A=1 which gives the A as

\displaystyle A=\frac{\pi}{4}+2n\pi where n is an integer, i.e.

\displaystyle \frac{x}{5}-\frac{\pi}{12}=\frac{\pi}{4}+2n\pi which gives us \displaystyle x=\frac{5\pi}{3}+10n\pi.


2 comments: