Differentiation (Part-38) | S N De

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GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS (PART-14)

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In the previous article, we have discussed few short ans type questions in GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS (PART-13). Today we'll continue solving few more problems.

5. Show that the three equations \(\sin^2\theta=\sin^2\alpha,\ \cos^2\theta=\cos^2\alpha\) and \(\tan^2\theta=\tan^2\alpha\) are all identical and the solution of each equation is \(n\pi \pm \alpha.\)

Sol. For \(\sin^2\theta=\sin^2\alpha:\)

\[\,\sin^2\theta=\sin^2\alpha \\ \Rightarrow 1-2\sin^2\theta=1-2\sin^2\alpha \\ \Rightarrow \cos2\theta=\cos2\alpha \\ \Rightarrow 2\theta=2n\pi \pm 2\alpha \\ \Rightarrow \theta=n\pi \pm \alpha,\,\, n\in \mathbb Z \rightarrow(1)\]

For \(\cos^2\theta=\cos^2\alpha:\)

\[\,\,\cos^2\theta=\cos^2\alpha \\ \Rightarrow 2\cos^2\theta-1=2\cos^2\alpha-1 \\ \Rightarrow \cos2\theta=\cos2\alpha \\ \Rightarrow 2\theta=2n\pi \pm 2\alpha \\ \Rightarrow \theta=n\pi \pm \alpha,\,\,n\in\mathbb Z \rightarrow(2)\]

For \(\tan^2\theta=\tan^2\alpha:\)

\[\,\tan^2\theta=\tan^2\alpha \\ \Rightarrow 1+\tan^2\theta=1+\tan^2\alpha \\ \Rightarrow \sec^2\theta=\sec^2\alpha \\ \Rightarrow \cos^2\theta=\cos^2\alpha \\ \Rightarrow 2\cos^2\theta-1=2\cos^2\alpha-1\\ \Rightarrow \cos2\theta=\cos2\alpha \\ \Rightarrow 2\theta=2n\pi \pm 2\alpha \\ \Rightarrow \theta=n\pi \pm \alpha,\,\,n\in\mathbb Z \rightarrow(3)\]

From \((1),(2),(3)\) we can say that the three equations \(\sin^2\theta=\sin^2\alpha,\ \cos^2\theta=\cos^2\alpha\) and \(\tan^2\theta=\tan^2\alpha\) are all identical and the solution of each equation is \(n\pi \pm \alpha.\)

6. Solve (general solution is not needed): \(\tan x+\tan y=2,\quad2\cos x\cos y=1.\)

Sol. We have,

\[2\cos x\cos y=1\rightarrow(1)\]

Now,

\[\,\tan x+\tan y=2 \\ \Rightarrow \frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}=2 \\ \Rightarrow \frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}=2 \\ \Rightarrow \sin(x+y)=2\cos x\cos y \\ \Rightarrow \sin(x+y)=1\,\,\,[\text{By (1)}] \\ \Rightarrow x+y=\pi/2 \rightarrow(2)\]

From \((1),(2)\) we get,

\[\,2\cos x \cos(\pi/2-x)=1 \\ \Rightarrow 2\cos x\sin x=1 \\ \Rightarrow \sin 2x=1=\sin(\pi/2) \\ \Rightarrow 2x=\pi/2 \\ \Rightarrow x=\pi/4 \rightarrow(3)\]

Solving \((2),(3)\) we get, \(x=y=\pi/4.\)

7. Find the least positive values of \(x\) and \(y\): \(\sin(x-y)=\frac 12,\quad\cos(x+y)=\frac 12.\)

Sol. We have,

\[\,\sin(x-y)=\frac 12=\sin(\pi/6)\\ \Rightarrow x-y=\frac{\pi}{6}\rightarrow(1)\]

Again,

\[\,\,\cos(x+y)=\frac 12=\cos(\pi/3) \\ \Rightarrow x+y=\frac{\pi}{3}\rightarrow(2)\]

Solving, \((1),(2)\) we get, \(x=\pi/4,\ y=\pi/12.\)

8. Find the least positive values of \(x\) and \(y\) so that \(2(\sin x+\cos x)-2\cos(x-y)=3.\)

Sol.

\[\,\,2(\sin x+\cos x)-2\cos(x-y)=3 \\ \Rightarrow 4\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ -2\left[2\cos^2\left(\frac{x-y}{2}\right)-1\right]-3=0 \\ \Rightarrow 4\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\-4\cos^2\left(\frac{x-y}{2}\right)-1=0 \\ \Rightarrow 4\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)-4\cos^2\left(\frac{x-y}{2}\right) \\-\sin^2\left(\frac{x+y}{2}\right)-\cos^2\left(\frac{x+y}{2}\right)=0 \\ \Rightarrow -\left[\sin^2\frac{x+y}{2}-4\sin\frac{x+y}{2}\cos\frac{x-y}{2}\\ +4\cos^2\frac{x-y}{2}\right]-\cos^2\frac{x+y}{2}=0 \\ \Rightarrow \left(\sin\frac{x+y}{2}-2\cos\frac{x-y}{2}\right)^2+\cos^2\frac{x+y}{2}=0 \\ \therefore \,\,\,\sin\frac{x+y}{2}-2\cos\frac{x-y}{2}=0\rightarrow(1), \\ \text{and}\,\,\, \cos\frac{x+y}{2}=0 \rightarrow(2)\]

From \((2)\) we get,

\[\,\frac{x+y}{2}=\pi/2 \\ \Rightarrow x+y=\pi\rightarrow(3)\]

By \((1),(3)\) we get,

\[\,\,\sin (\pi/2)-2\cos\frac{x-y}{2}=0 \\ \Rightarrow 2\cos\frac{x-y}{2}=1 \\ \Rightarrow \cos\frac{x-y}{2}=\frac 12 \\ \Rightarrow \frac{x-y}{2}=\pm\frac{\pi}{6} \\ \Rightarrow x-y=\frac{2\pi}{3},\\\text{or},\,\,y-x=\frac{2\pi}{3}\rightarrow(4)\]

Solving \((3),(4)\) we get, \(x=\frac{5\pi}{6},y=\frac{\pi}{6}\) or \(x=\pi/6,y=5\pi/6.\)

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