Differentiation (Part-38) | S N De

 

$\,\,2(i)\,\,$ Solve : $\,\,\cos(2x+3y)=\frac 12\rightarrow(1)\\ \cos(3x+2y)=\frac{\sqrt3}{2}\rightarrow(2)$

From $\,(1)\,$ we get, $\,\,\cos(2x+3y)=\frac 12=\cos\frac{\pi}{3}\\ \Rightarrow 2x+3y=2n\pi \pm\frac{\pi}{3} \rightarrow(3)$

Again, from $\,(2)\,$ we get, $\,\,\cos(3x+2y)=\frac{\sqrt3}{2}=\cos\frac{\pi}{6} \\ \Rightarrow 3x+2y=2m\pi\pm \frac{\pi}{6}\rightarrow(4)$

Here, $\,\,n,m\,$ are any integer.

Now, By $\,\,(4)\times 3-(3) \times 2,\,\,$ we get, 

$\,\,5x=3\left(2m\pi\pm \frac{\pi}{6}\right)-2\left(2n\pi \pm\frac{\pi}{3}\right) \\ \Rightarrow x=\frac 15\left[2(3m-2n)\pi \pm\frac{\pi}{2} \pm\frac{2\pi}{3}\right] \rightarrow(5)$

Similarly, by $\,\,(4) \times 2- (3) \times 3,\,\,$ we get,

$\,\,(6x+4y)-(6x+9y)=2\left(2m\pi \pm\frac{\pi}{6}\right)\\ -3\left(2n\pi \pm\frac{\pi}{3}\right) \\ \Rightarrow 5y=-2\left(2m\pi \pm\frac{\pi}{6}\right)+3\left(2n\pi \pm\frac{\pi}{3}\right)\\ \Rightarrow y=\frac 15\left[(3n-2m) \pm \pi \mp\frac{\pi}{3}\right]\rightarrow(6)$

Hence, from $\,\,(5)\, , (6)\,\,$ we get the required general solution. 

Solve :

$\,2(ii)\, \cos x+\cos y-\cos(x+y)\\=\frac 32  \,\,[0<x,y<\pi\,\,\text{(given)}]\\ \Rightarrow 2 \times 2 \cos\frac{x+y}{2} \cos\frac{x-y}{2}\\-2  \left(2\cos^2\frac{x+y}{2}-1\right)=3 \\ \Rightarrow 4\cos\frac{x+y}{2} \cos\frac{x-y}{2}-4\cos^2\frac{x+y}{2}\\+2=3 \\ \Rightarrow 4\cos^2\frac{x+y}{2}-4\cos\frac{x+y}{2}\\ \times  \cos\frac{x-y}{2}+3-2=0 \\ \Rightarrow \left(2\cos\frac{x+y}{2}-\cos\frac{x-y}{2}\right)^2\\+1-\cos^2\frac{x-y}{2}=0 \\ \Rightarrow \left(2\cos\frac{x+y}{2}-\cos\frac{x-y}{2}\right)^2\\+\sin^2\frac{x-y}{2}=0 \\ \therefore 2\cos\frac{x+y}{2}-\cos\frac{x-y}{2}=0 \rightarrow(1) \\ \text{and}\,\,\,\sin\frac{x-y}{2}=0 \rightarrow(2)$

From $\,\,(2)\,\,$ we get, $\,\,\cos\frac{x-y}{2}=1 \rightarrow(3)$

Again, from $\,\,(1),(3),\,\,$ we get, 

$\,\,\cos\frac{x+y}{2}=\frac 12=\cos\frac{\pi}{3} \\ \Rightarrow \frac{x+y}{2}=\frac{\pi}{3} \\ \Rightarrow x+y=\frac{2\pi}{3} \rightarrow(4)$

Again, from $\,(2)\,$ we get, $\,\,x-y=0 \Rightarrow x=y \rightarrow(5)$

Hence, solving  $\,\,(4), (5)\,\,$ we get, $\,\,x=y=\frac{\pi}{3}\,\,\text{(proved)}$

$\,\,3.\,\,$ Show that the following two equations represent the same set of angles : $\,\,\theta-\frac{\pi}{6}=2n\pi \pm \frac{\pi}{3}\,\,$ and $\,\,\theta+\frac{\pi}{3}=n\pi+(-1)^n\frac{\pi}{6}.$

Sol.  $\,\,\cos\left(\theta-\frac{\pi}{6}\right)=\cos\left(2n\pi \pm \frac{\pi}{3}\right) \\ \therefore  \sin \left[\frac{\pi}{2}-\left(\theta-\frac{\pi}{6}\right)\right]=\cos\frac{\pi}{3} \\ \Rightarrow \sin\left(\frac{2\pi}{3}-\theta\right)=\frac 12 \\ \Rightarrow \sin\left[\pi-\left(\frac{\pi}{3}+\theta\right)\right]=\frac 12 \\ \Rightarrow \sin\left(\theta+\frac{\pi}{3}\right)=\sin\frac{\pi}{6} \\ \Rightarrow \theta+\frac{\pi}{3}=n\pi+(-1)^n \frac{\pi}{6}\,\,\text{(showed)}$

$\,4(i)\,$ If $\,\,\sin\left(\frac{\pi}{2}\cos\theta\right)=\cos\left(\frac{\pi}{2} \sin\theta\right),\,\,$ show that $\,\,\pm \cos\left(\theta\mp \frac{\pi}{4}\right)=\frac{4n+1}{\sqrt2}\,\,\,$ where $\,n\,$ is any integer.

Sol. $\,\,\sin\left(\frac{\pi}{2}\cos\theta\right)=\cos\left(\frac{\pi}{2} \sin\theta\right) \\ \Rightarrow \cos\left(\frac{\pi}{2} \sin\theta\right)=\sin\left(\frac{\pi}{2}\cos\theta\right) \\ \Rightarrow \cos\left(\frac{\pi}{2} \sin\theta\right)=\cos\left(\frac{\pi}{2}-\frac{\pi}{2}\cos\theta\right) \\ \Rightarrow \frac{\pi}{2} \sin\theta=2n\pi \pm \left(\frac{\pi}{2}-\frac{\pi}{2}\cos\theta\right) \\ \Rightarrow \sin\theta=4n \pm (1-\cos\theta) \\ \Rightarrow \sin\theta \pm \cos\theta=4n \pm 1 \\ \Rightarrow \frac{1}{\sqrt2}\sin\theta \pm \frac{1}{\sqrt2}\cos\theta=\frac{4n \pm 1}{\sqrt2} \\ \therefore \pm \cos\left(\theta \mp \frac{\pi}{4}\right)=\frac{4n \pm 1}{\sqrt2}\,\,\,\text{(showed)}$

$\,4(ii)\,\,$  If $\,\,\tan(\pi\cos\theta)=\cot(\pi\sin\theta),\,\,$ prove that $\,\,\cos\left(\theta-\frac{\pi}{4}\right)=\frac{2n+1}{2\sqrt2}; n=0,\pm1,\pm2,\cdots$

Sol. We have, $\,\,\tan(\pi\cos\theta)=\cot(\pi\sin\theta)\\ \Rightarrow \tan (\pi \cos \theta)=\tan\left(\frac{\pi}{2}-\pi\sin\theta\right) \\ \Rightarrow \pi \cos\theta=n\pi+\frac{\pi}{2}-\pi \sin\theta \\ \Rightarrow \pi(\cos\theta+\sin\theta)=\frac{\pi}{2}(2n+1) \\ \Rightarrow \cos\theta+\sin\theta=\frac{2n+1}{2} \\ \Rightarrow \frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta=\frac{2n+1}{2\sqrt2} \\ \Rightarrow (\cos \frac{\pi}{4})\cos\theta+(\sin\frac{\pi}{4})\sin\theta=\frac{2n+1}{2\sqrt2} \\ \Rightarrow \cos\left(\theta-\frac{\pi}{4}\right)=\frac{2n+1}{2\sqrt2}, \quad n=0, \pm1,\pm2,\cdots$

$\,5.\,\,$ If $\,\csc A=\csc B\,\,$ and $\,\,\sec A=\sec B,\,\,$ then prove that, either $\,A=B\,$ or they differ by multiple of  four right angles. 

Sol. We have, $\,\,csc A=\csc B \\ \Rightarrow \sin A=\sin B \\ \Rightarrow \cos \left(\frac{\pi}{2}-A\right)=\cos \left(\frac{\pi}{2}-B\right) \\ \Rightarrow \cos \left(A-\frac{\pi}{2}\right)=\cos \left(\frac{\pi}{2}-B\right) \\ \Rightarrow A-\frac{\pi}{2}= 2n\pi\pm  \left(\frac{\pi}{2}-B\right) \rightarrow(1)$

Hence, from $\,(1),\,\,A-\pi/2=2n\pi +\pi/2-B \\ \Rightarrow A=(2n+1)\pi-B$

Again, by $\,(1)\,$, we get, $\,\,A-\pi/2=2n\pi -\pi/2+B \\ \Rightarrow A=2n\pi+B \rightarrow(2)$ 

Also, $\,\,\sec A=\sec B \\ \Rightarrow \cos A=\cos B \\ \Rightarrow A=2n\pi \pm B\rightarrow (3)$

From $\,(2)\,$ and $\,(3)\,$, we get , either $\,\,A=B\,\,$ which can be obtained by putting $\,\,n=0\,\,$ or they differ by multiple of  four right angles.