Differentiation (Part-38) | S N De

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): $\,\tan x+\tan y=2,\,\,2\cos x\cos y=1.$

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 $\,y :\,\, \sin(x-y)=\frac 12,\,\,\cos(x+y)=\frac 12.$

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 $\,\,x=\pi/6,\,\,y=5\pi/6.$