Differentiation (Part-38) | S N De

 $\,1(xxxviii)\,\,3\tan\left(\theta-\frac{\pi}{12}\right)\\=\tan\left(\theta+\frac{\pi}{12}\right)\rightarrow(1)$

Let $\,\,\theta-\frac{\pi}{12}=p,\,\,\theta+\frac{\pi}{12}=q \\ \text{so that}\,\, q+p=2\theta \rightarrow(2), \\ \text{and}\,\, q-p=2\pi/12=\pi/6\rightarrow(3)$

Then , from $\,(1)\,\,$ we get, $\,3\tan p=\tan q \\ \Rightarrow 3\frac{\sin p}{\cos p}=\frac{\sin q}{\cos q} \\ \Rightarrow 3=\frac{\sin q \cos p}{\cos q \sin p} \\ \Rightarrow \frac{3+1}{3-1}=\frac{\sin q\cos p+\cos q \sin p}{\sin q\cos p-\cos q\sin p} \\ \Rightarrow \frac{\sin (q+p)}{\sin(q-p)}=\frac 42 \\ \Rightarrow  \frac{\sin 2\theta}{\sin (\pi/6)}=2\,\,[\text{By (2),(3)}]\\ \Rightarrow  \sin 2\theta=2\sin(\pi/6)\\ \Rightarrow  \sin2\theta=2 \times \frac 12=1 \\ \Rightarrow  2\theta=(4n+1)\pi/2,\,\,n\in\mathbb Z  \\ \Rightarrow \theta=(4n+1)\pi/4=n\pi+\pi/4\, \,\text{(ans.)}$

$\,1(xxxix)\,\,\sec4\theta-\sec2\theta=2 \\ \Rightarrow \frac{1}{\cos4\theta}-\frac{1}{\cos2\theta}=2 \\ \Rightarrow \frac{\cos2\theta-\cos4\theta}{\cos4\theta\cos2\theta}=2 \\ \Rightarrow \cos2\theta-\cos4\theta=2\cos4\theta\cos2\theta \\ \Rightarrow \cos2\theta-\cos4\theta=\cos6\theta+\cos2\theta \\ \Rightarrow \cos6\theta+\cos4\theta=0 \\ \Rightarrow 2\cos\left(\frac{6\theta+4\theta}{2}\right) \cos\left(\frac{6\theta-4\theta}{2}\right)=0 \\ \Rightarrow 2\cos5\theta\cos\theta=0 \\ \therefore \cos5\theta=0 \rightarrow(1),\\ \text{or},\,\,\,\cos\theta=0\rightarrow(2).$

By $\,(1)\,\,$, we get, $\,\,5\theta=(2n+1)\pi/2,\,\,n\in\mathbb Z \\ \Rightarrow \theta=(2n+1)\pi/10\rightarrow(3)$

Again, by $\,(2)\,\,$, we get, $\,\,\theta=(2n+1)\pi/2\\ \Rightarrow \theta=(10n+5)\pi/10 \\ \Rightarrow \theta=[2(5n+2)+1]\frac{\pi}{10}\rightarrow(4)$

Hence, from $\,(3),\,(4)$, it follows that the general solution is given by : $\,\,\theta=(2n+1)\pi/10,\,\,n\in\mathbb Z.$

$\,1(xL)\,\,\sin x+\cos x=\frac{\cos 2x}{1-\sin2x} \\ \Rightarrow  \sin x+\cos x=\frac{\cos^2x-\sin^2x}{\cos^2x-2\sin x\cos x+\sin^2x} \\ \Rightarrow  \sin x+\cos x=\frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x-\sin x)^2} \\ \Rightarrow  \sin x+\cos x=\frac{\cos x+\sin x}{\cos x-\sin x} \\ \Rightarrow  (\sin x+\cos x)(\cos x-\sin x)\\ =(\cos x+\sin x) \\ \Rightarrow  (\sin x+\cos x)(\cos x-\sin x-1)=0 \\ \therefore \sin x+\cos x=0\rightarrow(1), \\ \text{or},\,\,\,\cos x-\sin x-1=0\rightarrow(2).$

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

 $\,\,\sin x=-\cos x \\ \Rightarrow \tan x=-1 \\ \Rightarrow x=n\pi-\frac{\pi}{4},\,\,n\in\mathbb Z.$

Again, by $\,(2),$ we get,

 $\,\frac{1}{\sqrt2}\cos x-\frac{1}{\sqrt2}\sin x=\frac{1}{\sqrt2}\\ \Rightarrow \cos\left(x+\frac{\pi}{4}\right)=\cos(\pi/4) \\ \Rightarrow  x+\pi/4=2n\pi \pm\pi/4 \\ \Rightarrow x=2n\pi,\,\, 2n\pi-\pi/2.$

Hence, the required general solution is given by : $\,\,x=n\pi-\pi/4,\,\,2n\pi,\,\,2n\pi-\pi/2,\,\,n\in\mathbb Z.$

$\,2.\,$ Find the general value of $\,\theta\,$ which satisfy $\,2\cos\theta+1=0,\,2\sin\theta-\sqrt3=0.$

Sol.  $\,\,2\cos\theta+1=0 \\ \Rightarrow \cos\theta=-\frac 12=\cos(2\pi/3) \\ \Rightarrow \theta=2n\pi \pm2\pi/3\rightarrow(1)$

Again, $\,\,2\sin\theta-\sqrt3=0 \\ \Rightarrow \sin\theta=\frac{\sqrt3}{2}=\sin(\pi/3) \\ \Rightarrow \theta=n\pi+(-1)^n \frac{\pi}{3} \rightarrow(2)$

Now, for $\,n\,$ even $\,=2k,\,$ say, we have from $\,(2)\,$, $\,\theta=2k\pi+\frac{\pi}{3} \rightarrow(3)$

Combining $\,(1),\,\,(3)$ we get, the general solution $\,\,\theta=2n\pi+2\pi/3\,\,n\in\mathbb Z.$

$\,3.\,$ If $\,\cos x+\sin x=\cos\alpha-\sin\alpha,\,\,$ prove that, $\,x-\pi/4=2n\pi \pm \left(\alpha+\pi/4\right)$

Sol. $\,\,\cos x+\sin x=\cos \alpha-\sin\alpha \\ \Rightarrow \cos x-\cos\alpha=-(\sin x+\sin\alpha) \\ \Rightarrow -2 \sin\left(\frac{x+\alpha}{2}\right)\sin\left(\frac{x-\alpha}{2}\right)=-2\sin\left(\frac{x+\alpha}{2}\right)\cos\left(\frac{x-\alpha}{2}\right) \\ \Rightarrow \sin\left(\frac{x+\alpha}{2}\right)\left[\sin\left(\frac{x-\alpha}{2}\right)-\cos\left(\frac{x-\alpha}{2}\right)\right]=0 \\ \therefore \,\, \sin\left(\frac{x+\alpha}{2}\right)=0\rightarrow(1),\\ \text{or},\,\,\sin\left(\frac{x-\alpha}{2}\right)-\cos\left(\frac{x-\alpha}{2}\right)=0 \rightarrow(2)$

From $\,(1)\,$ we get, $\,\frac{x+\alpha}{2}=n\pi \\ \Rightarrow x=2n\pi-\alpha \\ \Rightarrow x-\pi/4=2n\pi-(\alpha+\pi/4) \rightarrow(3)$

Again, by $\,(2)\,$ we get, $\,\,\sin\frac{x-\alpha}{2}=\cos\frac{x-\alpha}{2} \\ \Rightarrow \tan\left(\frac{x-\alpha}{2}\right)=1 \\ \Rightarrow \frac{x-\alpha}{2}=n\pi+\pi/4 \\ \Rightarrow x-\alpha=2n\pi+\pi/2 \\ \Rightarrow x-\pi/4=2n\pi+\pi/2-\pi/4+\alpha  \\ \Rightarrow x-\pi/4=2n\pi+(\alpha+\pi/4)\rightarrow(4)$

Hence, from $\,(3),\,(4)\,\,$ we get, $\,x-\pi/4=2n\pi \pm(\alpha+\pi/4)\,\,\text{(proved)}$

Alternative Method :

$\,\,\cos x+\sin x= \cos\alpha-\sin\alpha \\ \Rightarrow \frac{1}{\sqrt2}\cos x+\frac{1}{\sqrt2} \sin x=\frac{1}{\sqrt2}\cos\alpha-\frac{1}{\sqrt2}\sin\alpha \\ \Rightarrow \cos\left(x-\frac{\pi}{4}\right)=\cos\left(\alpha+\frac{\pi}{4}\right) \\ \therefore \,\,x-\frac{\pi}{4}=2n\pi\pm \left(\alpha+\pi/4\right),\,\,n\in\mathbb Z.$

$\,4.\,$ If $\,\,\tan ax=\tan bx\,\,(a \neq b),\,\,$ show that , the values of $\,x\,$ form an  A.P.

Sol. $\,\tan ax=\tan bx \\ \Rightarrow ax=n\pi+bx \\ \Rightarrow x(a-b)=n\pi \\ \Rightarrow x=\frac{n\pi}{a-b},\,\,n\in\mathbb Z\rightarrow(1)$

Hence , from $\,(1)\,$ we get, the values of $\,x\,$ are given by $\cdots,\,\,\frac{-3\pi}{a-b},\,\frac{-2\pi}{a-b},\,\,\frac{-\pi}{a-b},\,0,\,\frac{\pi}{a-b},\,\frac{2\pi}{a-b},\,\,\frac{3\pi}{a-b},\cdots$ which clearly are in A.P. with common difference  $\,\,\frac{\pi}{a-b}.$

To continue with GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS (PART-14), click here.

Read More : 

• COMPLEX NUMBERS
• QUADRATIC EQUATIONS
• TRIGONOMETRIC RATIOS OF MULTIPLE ANGLES
• RELATION AND MAPPING
• TRIGONOMETRIC RATIOS OF SUBMULTIPLE ANGLES
• TRIGONOMETRIC RATIOS OF COMPOUND ANGLES