Question
Prove that:
$\sin5\text{x}=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}$
$\sin5\text{x}=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column A | Column B | ||
| a. | The polar form of $\text{i}+\sqrt{3}$ is | i. | Perpendicular bisector of segment joining (– 2, 0) and (2, 0). |
| b. | The amplitude of $-1+\sqrt{-3}$ is | ii. | On or outside the circle having centre at (0, – 4) and radius 3. |
| c. | If |z + 2| = |z - 2|, then locus of z is | iii. | $\frac{2\pi}{3}$ |
| d. | If |z + 2i| = |z - 2i|, then locus of z is | iv. | Perpendicular bisector of segment joining (0, – 2) and (0, 2). |
| e. | Region represented by $|\text{z}+4\text{i}|\geq3$ is | v. | $2\Big(\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$ |
| f. | Region represented by $|\text{z}+4|\leq3$ is | vi. | On or inside the circle having centre (– 4, 0) and radius 3 units. |
| g. | Conjugate of $\frac{1+2\text{i}}{1-\text{i}}$ lies in | vii. | First quadrant. |
| h. | Reciprocal of 1 - i lies in | viii. | Third quadrant. |