MCQ
$2(1-2\sin^27\text{x})\sin3\text{x}$ is equal to:
- A$\sin17\text{x}-\sin11\text{x}$
- B$\sin11\text{x}-\sin17\text{x}$
- C$\cos17\text{x}-\cos11\text{x}$
- D$\cos17\text{x}+\cos11\text{x}$
Solution:
We have,
$2(1-2\sin^27\text{x})\sin3\text{x}=2(\cos14\text{x})\sin3\text{x}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=2\sin3\text{x}\cos14\text{x}$
$=\sin17\text{x}-\sin11\text{x}$ $[\because2\sin\text{A}\cos\text{xB}=\sin(\text{A+B})-\sin(\text{A}-\text{B})]$
$\therefore2(1-2\sin^27\text{x})\sin3\text{x}=\sin17\text{x}-\sin11\text{x}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in: