Class 11: Trigonometric Functions – Very Short Answer Questions

$\displaystyle \textbf{Question 1. }\text{Write the maximum and minimum values of }\cos(\cos x).$
$\displaystyle \text{Answer:}$
$\displaystyle -1\leq \cos x\leq1$
$\displaystyle \therefore \cos(\cos x)\text{ lies between }\cos1\text{ and }1$
$\displaystyle \text{Maximum value}=1$
$\displaystyle \text{Minimum value}=\cos1$
$\\$

$\displaystyle \textbf{Question 2. }\text{Write the maximum and minimum values of }\sin(\sin x).$
$\displaystyle \text{Answer:}$
$\displaystyle -1\leq \sin x\leq1$
$\displaystyle \therefore -\sin1\leq \sin(\sin x)\leq\sin1$
$\displaystyle \text{Maximum value}=\sin1$
$\displaystyle \text{Minimum value}=-\sin1$
$\\$

$\displaystyle \textbf{Question 3. }\text{Write the maximum value of }\sin(\cos x).$
$\displaystyle \text{Answer:}$
$\displaystyle -1\leq \cos x\leq1$
$\displaystyle \therefore \sin(\cos x)\leq\sin1$
$\displaystyle \text{Maximum value}=\sin1$
$\\$

$\displaystyle \textbf{Question 4. }\text{If }\sin x=\cos^2x,\text{ then write the value of }\cos^2x(1+\cos^2x).$
$\displaystyle \text{Answer:}$
$\displaystyle \sin x=\cos^2x$
$\displaystyle \therefore \cos^2x=\sin x$
$\displaystyle \cos^2x(1+\cos^2x)=\sin x(1+\sin x)$
$\displaystyle =\sin x+\sin^2x$
$\displaystyle =\sin x+(1-\cos^2x)$
$\displaystyle =\sin x+1-\sin x=1$
$\\$

$\displaystyle \textbf{Question 5. }\text{If }\sin x+\mathrm{cosec}\ x=2,\text{ then write the value of }\sin^nx+\mathrm{cosec}^nx.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin x+\frac{1}{\sin x}=2$
$\displaystyle \therefore \sin x=1$
$\displaystyle \therefore \mathrm{cosec}\ x=1$
$\displaystyle \therefore \sin^nx+\mathrm{cosec}^nx=1^n+1^n=2$
$\\$

$\displaystyle \textbf{Question 6. }\text{If }\sin x+\sin^2x=1,\text{ then write the value of } \\ \cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin x+\sin^2x=1$
$\displaystyle \therefore 1-\sin^2x=\sin x$
$\displaystyle \therefore \cos^2x=\sin x$
$\displaystyle \cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x=\cos^6x(\cos^2x+1)^3$
$\displaystyle =(\cos^2x)^3(1+\cos^2x)^3$
$\displaystyle =\{\sin x(1+\sin x)\}^3$
$\displaystyle =1^3=1$
$\\$

$\displaystyle \textbf{Question 7. }\text{If }\sin x+\sin^2x=1,\text{ then write the value of } \\ \cos^8x+2\cos^6x+\cos^4x.$
$\displaystyle \text{Answer:}$
$\displaystyle \cos^8x+2\cos^6x+\cos^4x=\cos^4x(\cos^2x+1)^2$
$\displaystyle =(\cos^2x)^2(1+\cos^2x)^2$
$\displaystyle =\{\sin x(1+\sin x)\}^2$
$\displaystyle =1^2=1$
$\\$

$\displaystyle \textbf{Question 8. }\text{If }\sin\theta_1+\sin\theta_2+\sin\theta_3=3,\text{ then write the value of } \\ \cos\theta_1+\cos\theta_2+\cos\theta_3.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin\theta_1+\sin\theta_2+\sin\theta_3=3$
$\displaystyle \therefore \sin\theta_1=\sin\theta_2=\sin\theta_3=1$
$\displaystyle \therefore \cos\theta_1=\cos\theta_2=\cos\theta_3=0$
$\displaystyle \therefore \cos\theta_1+\cos\theta_2+\cos\theta_3=0$
$\\$

$\displaystyle \textbf{Question 9. }\text{Write the value of }\sin10^\circ+\sin20^\circ+\sin30^\circ+\cdots+\sin360^\circ.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin\theta+\sin(360^\circ-\theta)=0$
$\displaystyle \text{All terms cancel in pairs, except }\sin90^\circ\text{ and }\sin270^\circ$
$\displaystyle \sin90^\circ+\sin270^\circ=1-1=0$
$\displaystyle \therefore \text{Required value}=0$
$\\$

$\displaystyle \textbf{Question 10. }\text{A circular wire of radius }15\text{ cm is cut and bent so as to lie along the} \\ \text{circumference of a loop of radius }120\text{ cm. Write the measure of the angle subtended} \\ \text{by it at the centre of the loop.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of wire}=2\pi(15)=30\pi\text{ cm}$
$\displaystyle \text{Radius of loop}=120\text{ cm}$
$\displaystyle \theta=\frac{\text{arc length}}{\text{radius}}=\frac{30\pi}{120}=\frac{\pi}{4}$
$\displaystyle \therefore \theta=45^\circ$
$\\$

$\displaystyle \textbf{Question 11. }\text{Write the value of }2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)+1.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
$\displaystyle \sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta$
$\displaystyle 2(1-3\sin^2\theta\cos^2\theta)-3(1-2\sin^2\theta\cos^2\theta)+1=0$
$\displaystyle \therefore \text{Required value}=0$
$\\$

$\displaystyle \textbf{Question 12. }\text{Write the value of }\cos1^\circ+\cos2^\circ+\cos3^\circ+\cdots+\cos180^\circ.$
$\displaystyle \text{Answer:}$
$\displaystyle \cos\theta+\cos(180^\circ-\theta)=0$
$\displaystyle \cos90^\circ=0,\quad \cos180^\circ=-1$
$\displaystyle \therefore \text{Required value}=0$
$\\$

$\displaystyle \textbf{Question 13. }\text{If }\cot(\alpha+\beta)=0,\text{ then write the value of }\sin(\alpha+2\beta).$
$\displaystyle \text{Answer:}$
$\displaystyle \cot(\alpha+\beta)=0$
$\displaystyle \therefore \alpha+\beta=\frac{\pi}{2}$
$\displaystyle \sin(\alpha+2\beta)=\sin\left(\frac{\pi}{2}+\beta\right)$
$\displaystyle =\cos\beta$
$\\$

$\displaystyle \textbf{Question 14. }\text{If }\tan A+\cot A=4,\text{ then write the value of }\tan^4A+\cot^4A.$
$\displaystyle \text{Answer:}$
$\displaystyle \tan A+\cot A=4$
$\displaystyle \tan^2A+\cot^2A+2=16$
$\displaystyle \tan^2A+\cot^2A=14$
$\displaystyle \tan^4A+\cot^4A+2=196$
$\displaystyle \tan^4A+\cot^4A=194$
$\\$

$\displaystyle \textbf{Question 15. }\text{Write the least value of }\cos^2\theta+\sec^2\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \cos^2\theta+\sec^2\theta=\cos^2\theta+\frac{1}{\cos^2\theta}$
$\displaystyle a+\frac{1}{a}\geq2$
$\displaystyle \therefore \text{Least value}=2$
$\\$

$\displaystyle \textbf{Question 16. }\text{If }x=\sin^{14}\theta+\cos^{20}\theta,\text{ then write the smallest interval in} \\ \text{which the value of }x\text{ lie.}$
$\displaystyle \text{Answer:}$
$\displaystyle 0\leq\sin^{14}\theta\leq1,\quad 0\leq\cos^{20}\theta\leq1$
$\displaystyle \sin^{14}\theta+\cos^{20}\theta\leq1$
$\displaystyle \therefore 0<x\leq1$
$\displaystyle \therefore x\in(0,1]$
$\\$

$\displaystyle \textbf{Question 17. }\text{If }3\sin\theta+5\cos\theta=5,\text{ then write the value of }5\sin\theta-3\cos\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle (3\sin\theta+5\cos\theta)^2+(5\sin\theta-3\cos\theta)^2=34$
$\displaystyle 25+(5\sin\theta-3\cos\theta)^2=34$
$\displaystyle (5\sin\theta-3\cos\theta)^2=9$
$\displaystyle \therefore 5\sin\theta-3\cos\theta=\pm3$