- ✓$\cos 2\theta $
- B$cos 3\theta$
- C$\sin 2\theta $
- D$\sin 3\theta $
Now, put $\theta = \phi = \frac{\pi }{4}$
$\cos 2\left( {\frac{\pi }{2}} \right) - 4\cos \left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{4}} \right) + 2{\sin ^2}\left( {\frac{{2\pi }}{4}} \right) = 0$
Put $\theta = \phi = \pi /4$ in option $(a)$,
then, $\cos 2\theta = \cos \pi /2 = 0$.
Hence option $(a)$ is correct.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$ A=\{z: \operatorname{Im} z \geq 1\} $
$ B=\{z:|z-2-i|=3\} $
$ C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\} .$
$1.$ The number of elements in the set $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}$ is
$(A)$ $0$ $(B)$ $1$ $(C)$ $2$ $(D)$ $\infty$
$2.$ Let $z$ be any point in $A \cap B \cap C$. Then, $|z+1-i|^2+|z-5-i|^2$ lies between
$(A)$ $25$ and $29$ $(B)$ $30$ and $34$ $(C)$ $35$ and $39$ $(D)$ $40$ and $44$
$3.$ Let $z$ be any point in $A \cap B \cap C$ and let $w$ be any point satisfying $|w-2-i|<3$. Then, $|z|-|w|+3$ lies between
$(A)$ $-6$ and $3$ $(B)$ $-3$ and $6$
$(C)$ $-6$ and $6$ $(D)$ $-3$ and $9$
Give the answer question $1,2$ and $3.$