Question
$\sin5\text{x}=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$
✓
Answer
$\text{LHS}=\sin5\text{x}$
$=\sin\text{x}(3\text{x}+2\text{x})$
$=\sin3\text{x}\times\cos2\text{x}+\cos3\text{x}\times\sin2\text{x}$
$=(3\sin\text{x}-4\sin^3\text{x})(2\cos^2-1)\\+(4\cos^32\cos^2-3\cos\text{x})\times2\sin\text{x}\cos\text{x}$
$=-3\sin\text{x}-4\sin^3\text{x}+6\sin\text{x}\cos^2\text{x}\\-8\sin^3\text{x}\cos^2\text{x}+8\sin\text{x}\cos^4\text{x}-6\sin\text{x}\cos^2\text{x}$
$=8\sin\text{x}\cos^4\text{x}-8\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}+4\sin^3\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}\\+3\sin\text{x}\cos^4\text{x}+4\sin^3\text{x}+2\sin^3\text{x}\cos^2\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^4\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^2\text{x})(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3(1+\cos^2\text{x})-2(2+\cos^2\text{x})]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3+3\cos^2\text{x}-4-2\cos^2\text{x}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}[\cos^2\text{x-1}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}\times(-\sin^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}+\sin^5\text{x}$
$=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$
$=\text{RHS}$
$=\sin\text{x}(3\text{x}+2\text{x})$
$=\sin3\text{x}\times\cos2\text{x}+\cos3\text{x}\times\sin2\text{x}$
$=(3\sin\text{x}-4\sin^3\text{x})(2\cos^2-1)\\+(4\cos^32\cos^2-3\cos\text{x})\times2\sin\text{x}\cos\text{x}$
$=-3\sin\text{x}-4\sin^3\text{x}+6\sin\text{x}\cos^2\text{x}\\-8\sin^3\text{x}\cos^2\text{x}+8\sin\text{x}\cos^4\text{x}-6\sin\text{x}\cos^2\text{x}$
$=8\sin\text{x}\cos^4\text{x}-8\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}+4\sin^3\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}\\+3\sin\text{x}\cos^4\text{x}+4\sin^3\text{x}+2\sin^3\text{x}\cos^2\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^4\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^2\text{x})(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3(1+\cos^2\text{x})-2(2+\cos^2\text{x})]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3+3\cos^2\text{x}-4-2\cos^2\text{x}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}[\cos^2\text{x-1}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}\times(-\sin^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}+\sin^5\text{x}$
$=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$
$=\text{RHS}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Calculate the mean, variance and standard deviation of the following frequency distribution.
| Class: | 1-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency: | 11 | 29 | 18 | 4 | 5 | 3 |
Calculate the standard deviation for the following data:
|
Class:
|
0-30 | 30-60 |
60-90
|
90-120
|
120-150 | 150-180 |
180-210
|
| Frequency: |
9
|
17
|
43
|
82
|
81 | 44 |
24
|
If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that
$\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
→$\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
Find the general solution for each of the following equations:
$\sin\text{x}+\sin3\text{x}+\sin5\text{x}=0$
→$\sin\text{x}+\sin3\text{x}+\sin5\text{x}=0$
Find the equation of the set of all points the sum of whose distances from the points $(3, 0)$ and $(9, 0)$ is $12.$
→Sum the following series to n terms:
$1 + 3 + 6 + 10 + 15 + .....$
→$1 + 3 + 6 + 10 + 15 + .....$
Determine the probability $p,$ for each of the following events.
→- An odd number appears in a single toss of a fair die.
- At least one head appears in two tosses of a fair coin.
- A king, $9$ of hearts, or $3$ of spades appears in drawing a single card from a well shuffled ordinary deck of $52$ cards.
- The sum of $6$ appears in a single toss of a pair of fair dice.
The accompanying Venn diagram shows three events, $A, B,$ and $C,$ and also the probabilities of the various intersections $($for instance, $\text{P}(\text{A}\cap\text{B})=.07)$ Determine:
→
- $\text{P(A)}$
- $\text{P}(\text{B}\cap\bar{\text{C}})$
- $\text{P(A}\cup\text{B})$
- $\text{P(A}\cap\bar{\text{B}})$
- $\text{P(B}\cap\text{C})$
- Probability of exactly one of the three occurs.
$\text{If}\ \cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D}),$
prove that $\tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$
→prove that $\tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$
One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes $S = \{$John promoted, Rita promoted, Aslam promoted, Gurpreet promoted$\}$ You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John:
$P($Rita promoted$)$
$P($Aslam promoted$)$
$P($Gurpreet promoted$)$
→- Determine $P($John promoted$)$
$P($Rita promoted$)$
$P($Aslam promoted$)$
$P($Gurpreet promoted$)$
- If $A = \{$John promoted or Gurpreet promoted$\}$, find $P(A).$