The minute hand of a watch is $1.5 \,cm$ long. How far does its tip move in $40$ minutes? ( Use $\pi=3.14$ ).
In $60$ minutes, the minute hand of a watch completes one revolution. Therefore, in $40$ minutes, the minute hand turns through $\frac{2}{3}$ of a revolution. Therefore, ${\theta = 23 \times {{360}^\circ }}$ or $\frac{4 \pi}{3}$ radian. Hence, the required distance travelled is given by
$l=r \theta=1.5 \times \frac{4 \pi}{3} \,cm =2 \pi \,cm =2 \times 3.14 \,cm =6.28 \,cm$
Let the function $:(0, \pi) \rightarrow R$ be defined by
$f (\theta)=(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .
Prove that:
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
The product $\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right)\left(1+\tan 3^{\circ}\right)$ $. .\left(1+\tan 45^{\circ}\right)$ equals
Prove that:
$ 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
Find the radian measures corresponding to the following degree measures:
$240^{\circ}$