MCQ
The value of $2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\cos^3\text{x}\sin^2\text{x}$ is:
- A$2$
- B$1$
- ✓$0$
- D$-1$
✓
Answer
Correct option: C.
$0$
We have,
$2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\cos^3\text{x}\sin^2\text{x}$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\Big[\frac{\cos3\text{x}+3\cos\text{x}}{4}\times\frac{(1-\cos2\text{x})}{2}\Big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[(\cos3\text{x}+3\cos\text{x})(1-\cos2\text{x})\big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[\cos3\text{x}-\cos3\text{x}\cos2\text{x}+3\cos\text{x}\cos2\text{x}\big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]\\+2\cos3\text{x}\cos2\text{x}+3[2\cos\text{x}\cos2\text{x}]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]+\cos5\text{x}+\cos\text{x}\\\ \ \ +3\cos3\text{x}+3\cos\text{x}[2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$
$2\cos\text{x}=\cos3\text{x}-\cos5\text{x}-2\cos3\text{x}-6\cos\text{x}\\+\cos5\text{x}+\cos\text{x}+3\cos3\text{x}+3\cos\text{x}=0$
$2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\cos^3\text{x}\sin^2\text{x}$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\Big[\frac{\cos3\text{x}+3\cos\text{x}}{4}\times\frac{(1-\cos2\text{x})}{2}\Big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[(\cos3\text{x}+3\cos\text{x})(1-\cos2\text{x})\big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[\cos3\text{x}-\cos3\text{x}\cos2\text{x}+3\cos\text{x}\cos2\text{x}\big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]\\+2\cos3\text{x}\cos2\text{x}+3[2\cos\text{x}\cos2\text{x}]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]+\cos5\text{x}+\cos\text{x}\\\ \ \ +3\cos3\text{x}+3\cos\text{x}[2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$
$2\cos\text{x}=\cos3\text{x}-\cos5\text{x}-2\cos3\text{x}-6\cos\text{x}\\+\cos5\text{x}+\cos\text{x}+3\cos3\text{x}+3\cos\text{x}=0$
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