Differentiation (Part-38) | S N De

 

$\,1.\,$ If $\,\sin\theta=-\frac 35,\,\,180^{\circ}<\theta<270^{\circ},\,\,$ find the values of $\,\sin\frac{\theta}{2}\,\,$ and $\,\,\cos\frac{\theta}{2}.$

Sol. $\,\quad\sin\theta=-\frac 35 \\ \therefore\,\,\cos\theta \\=\pm\sqrt{1-\sin^2\theta}\\=\pm\sqrt{1-(-3/5)^2}\\=\pm\sqrt{1-9/25}\\=\pm\sqrt{\frac{16}{25}}\\=\pm\frac 45$

Since $\,\,180^{\circ}<\theta<270^{\circ},\,\, \cos\theta=-\frac 45 \\ \Rightarrow2\cos^2\theta-1=-\frac 45 \\ \Rightarrow \cos^2\frac{\theta}{2}=\frac{1}{10} \\ \therefore \cos\frac{\theta}{2}=\pm\frac{1}{\sqrt{10}}$

Since $\,\,180^{\circ}<\theta<270^{\circ}  \\ \therefore 90^{\circ}<\frac{\theta}{2}<135^{\circ}$

and so, $\,\frac{\theta}{2}\,\,$ lies in the 2nd quadrant and $\,\cos\frac{\theta}{2}\,\,$ is negative in that quadrant and $\,\sin\frac{\theta}{2}\,$ is positive in that quadrant $\rightarrow(1).$

Hence, $\,\,\cos\frac{\theta}{2}=-\frac{1}{\sqrt{10}} \\ \text{and}\,\,\sin\frac{\theta}{2}=+\sqrt{1-\cos^2\frac{\theta}{2}}\,\,[\text{By (1)}]\\~~~~~~~~~~~~~~~~~=\sqrt{1-\frac{1}{10}}\\~~~~~~~~~~~~~~~~~=\sqrt{\frac{9}{10}}\\~~~~~~~~~~~~~~~~~=\frac{3}{\sqrt{10}}$ 

$\,2(i)\,\,$ If $\,\,\theta\,$ lies in the fourth quadrant and $\,\,\sin\theta=-\frac{5}{13},\,\,$ find the values of $\,\sin\frac{\theta}{2},\,\cos\frac{\theta}{2}\,\,\text{and}\,\,\tan\frac{\theta}{2}.$

Sol. $\,\,\sin\theta=-\frac{5}{13} \\ \therefore \cos\theta=\pm\sqrt{1-(-5/13)^2}\\=\sqrt{1-\frac{25}{169}}\\=\pm\frac{12}{13}$

Since, $\,\,\theta\,$ lies in the fourth quadrant , so $\,\,\cos\theta=\frac{12}{13}.$ 

Now, $\,\,\cos\theta=\frac{12}{13} \\ \Rightarrow 2\cos^2\frac{\theta}{2}-1=\frac{12}{13} \\ \Rightarrow \cos^2\frac{\theta}{2}=\frac{25}{26} \\ \therefore \cos\frac{\theta}{2}=\pm \frac{5}{\sqrt{26}}$

By question, we have $\,\,\frac{3\pi}{2}<\theta<2\pi \\ \Rightarrow \frac{3\pi}{4}<\frac{\theta}{2}<\pi \\ \therefore \frac{\theta}{2}\,\,\text{lies in the 2nd quadrant and so}\,\,\cos\frac{\theta}{2}=-\frac{5}{\sqrt{26}}$

Again, $\,\,\sin\frac{\theta}{2}=\pm\sqrt{1-\cos^2\frac{\theta}{2}}\\=+\sqrt{1-\frac{25}{26}}\,\,[\text{Since},\,\,\frac{\theta}{2}\\\text{lies in the 2nd quadrant}]\\=\frac{1}{\sqrt{26}} \\ \therefore \tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}=-\frac 15$

$\,2(ii)\,$  If $\,\theta\,$ lies in the fourth quadrant and $\,\cos\theta=\frac 37,\,$ find the values of $\,\,\sin\frac{\theta}{2}\,\,$ and $\,\,\cos\frac{\theta}{2}.$

Sol. $\,\,\cos\theta=\frac 37 \\ \Rightarrow 2\cos^2\frac{\theta}{2}-1=\frac 37 \\ \Rightarrow \cos^2\frac{\theta}{2}=\frac{10}{14}=\frac 57 \\ \text{Since,}\,\,\,\frac{3\pi}{2}<\theta<2\pi \\ \Rightarrow \frac{3\pi}{4}<\frac{\theta}{2}<\pi$ 

Therefore, $\,\frac{\theta}{2}\,$ lies in the 2nd quadrant and in 2nd quadrant $\,\,\cos\frac{\theta}{2}\,\,$ is negative and $\,\,\sin\frac{\theta}{2}\,\,$ is positive.

So, $\,\,\cos\frac{\theta}{2}=-\sqrt{\frac 57} \\ \text{and}\,\,\sin\frac{\theta}{2}=\sqrt{1-\cos^2\frac{\theta}{2}}=\sqrt{1-\frac 57}=\sqrt{\frac 27}$

$\,2(iii)\,$  If $\,\theta\,$ lies in the third quadrant and $\,\tan\theta=\frac{5}{12},\,$ then find the values of $\,\sin\frac{\theta}{2}\,\,$ and $\,\,\cos\frac{\theta}{2}.$

Sol.  $\,\tan\theta=\frac{5}{12} \\ \Rightarrow \sec^2\theta=1+\tan^2\theta\\~~~~~~~~~~~~~~=1+\left(\frac{5}{12}\right)^2=1+\frac{25}{144} \\ \Rightarrow \sec^2\theta=\frac{144+25}{144}=\frac{169}{144} \\ \Rightarrow \cos^2\theta=\frac{144}{169} \\ \therefore \cos\theta=\pm\sqrt{\frac{144}{169}}=\pm\frac{12}{13}.$

Now, since $\,\theta\,$ lies in the third quadrant , $\,\,\cos\theta<0\,$ and so $\,\cos\theta=-\frac{12}{13}$

So, $\,\,\cos\theta=-\frac{12}{13} \\ \Rightarrow 2\cos^2\frac{\theta}{2}-1=-\frac{12}{13} \\ \Rightarrow 2\cos^2\frac{\theta}{2}=1-\frac{12}{13} \\ \Rightarrow \cos^2\frac{\theta}{2}=\frac{1}{26} \\ \Rightarrow\cos\frac{\theta}{2}=\pm \frac{1}{\sqrt{26}}$

Since, $\,\pi<\theta<\frac{3\pi}{2} \\ \Rightarrow \frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}$

Now, as $\,\frac{\theta}{2}\,$ lies in the 2nd quadrant, so $\,\cos\frac{\theta}{2}<0,\,\sin\frac{\theta}{2}>0.$

$\,\therefore \cos\frac{\theta}{2}=-\frac{1}{\sqrt{26}} \\ \text{and}\,\,\sin\frac{\theta}{2}=\sqrt{1-\frac{1}{26}}=\frac{5}{\sqrt{26}}$

$\,3.\,$ If $\,\theta=330^{\circ},\,$ then using the formula $\,\,\tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\,\,$ find the value of $\,\tan\frac{\theta}{2}.$

Sol. We put $\,\,\theta=330^{\circ}\,$ in the formula $\,\,\tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\,\,$ so that $\,\tan330^{\circ}=\frac{2x}{1-x^2},\rightarrow(1)\,\,\text{where}\,\,x=\tan\frac{\theta}{2}$

Now, by (1) we get $\,(1-x^2)\tan330^{\circ}=2x \\ \Rightarrow(1-x^2)\tan(4 \times 90^{\circ}-30^{\circ})=2x \\ \Rightarrow(1-x^2)(-\tan30^{\circ})=2x\\ \Rightarrow(1-x^2)(-\frac{1}{\sqrt3})=2x \\ \Rightarrow x^2-2\sqrt3x-1=0 \\ \Rightarrow x=\frac{2\sqrt3 \pm\sqrt{12+4}}{2}\\~~~~~~~~=\frac{2\sqrt3\pm4}{2}\\~~~~~~~~=\sqrt3\pm2\rightarrow(2)$

Since, $\,\,\theta=330^{\circ} \Rightarrow \theta=\frac{\theta}{2}=165^{\circ}\,$ which lies in 2nd quadrant and so in 2nd quadrant $\,\,\tan\frac{\theta}{2}<0\,$ which means from (2) we get $\,x=\sqrt3-2 \Rightarrow \tan\frac{\theta}{2}=\sqrt3-2.$

$\,4(i)\,$ Prove that $\,\,\frac{1+\sin\theta}{1-\sin\theta}=\tan^2(\frac{\pi}{4}+\frac{\theta}{2})$

Sol. $\,\,\frac{1+\sin\theta}{1-\sin\theta}\\=\frac{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)^2}{\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}\\=\left(\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}\right)^2\,\,[**]\\=\left(\frac{\tan\frac{\pi}{4}+\tan\frac{\theta}{2}}{1-\tan\frac{\pi}{4}\tan\frac{\theta}{2}}\right)^2\\=\left[\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right]^2\\=\tan^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)$

Note[**] : Dividing numerator and denominator by $\,\,\cos^2\frac{\theta}{2}$

$\,4(ii)\,$ Prove that $\,\,\tan\frac{A+B}{2}+\tan\frac{A-B}{2}=\frac{2\sin A}{\cos A+\cos B}$

Sol. $\,\,\tan\frac{A+B}{2}+\tan\frac{A-B}{2}\\=\frac{\sin\frac{A+B}{2}}{\cos\frac{A+B}{2}}+\frac{\sin\frac{A-B}{2}}{\cos\frac{A-B}{2}}\\=\frac{\sin\frac{A+B}{2}\cos\frac{A-B}{2}+\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{\cos\frac{A+B}{2}\cos\frac{A-B}{2}}\\=\frac{\sin\left(\frac{A+B}{2}+\frac{A-B}{2}\right)}{\frac 12\times \left[\cos\left(\frac{A+B}{2}+\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}-\frac{A-B}{2}\right)\right]}\\=\frac{2\sin A}{\cos A +\cos B}\,\,\text{(proved)}$

$\,4(iii)\,$ Prove that $\,\,\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\tan\frac{\alpha}{2}}{1-\tan\frac{\alpha}{2}}$

Sol. $\,\,\frac{\cos\alpha}{1-\sin\alpha}\\=\frac{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}-2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}\\=\frac{(\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2})(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2})}{(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2})^2}\\=\frac{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}\\=\frac{1-\tan\frac{\alpha}{2}}{1-\tan\frac{\alpha}{2}}\,\,[**]$

Note [**] : Dividing numerator and denominator by $\,\,\cos\frac{\alpha}{2}$

$\,4(iv)\,$ Prove that $\,\,(1+\cot\theta+\csc\theta)(1+\cot\theta-\csc\theta)=\cot\frac{\theta}{2}-\tan\frac{\theta}{2}$

Note : By $\csc \theta\,\,$ we mean $\,\,\text{cosec}\,\,\theta.$

Sol.  $\,\,(1+\cot\theta+\csc\theta)(1+\cot\theta-\csc\theta)\\=\frac{\sin\theta+\cos\theta-1}{\sin\theta} \times \frac{\sin\theta+\cos\theta+1}{\sin\theta}\\=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \times \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{2\sin\frac{\theta}{2}(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \times \frac{2\cos\frac{\theta}{2}(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\times \frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}}\\=\frac{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\cot\frac{\theta}{2}-\tan\frac{\theta}{2}\,\,\text{(proved)}$