English

Sin 5 X = 5 Cos 4 X Sin X − 10 Cos 2 X Sin 3 X + Sin 5 X

Advertisements
Advertisements

Question

\[\sin 5x = 5 \cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x\]

 

Numerical
Advertisements

Solution

\[LHS = \sin5x\]
\[ = \sin\left( 3x + 2x \right)\]
\[ = \sin3x \times \cos2x + \cos3x \times \sin2x\]
\[ = \left( 3\text{ sin } x - 4 \sin^3 x \right)\left( 2 \cos^2 x - 1 \right) + \left( 4 \cos^3 x - 3\text{ cos } x \right) \times 2\text{  sin } x \text{  cos } x\]
\[ = - 3\text{  sin } x + 4 \sin^3 x + 6\text{  sin } x \cos^2 x - 8 \sin^3 x \cos^2 x + 8\text{ sin } x \cos^4 x - 6\text{  sin } x \cos^2 x\]
\[ = 8\text{ sin } x \cos^4 x - 8 \sin^3 x \cos^2 x - 3\text{ sin }  x + 4 \sin^3 x\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3\text{ sin } x + 3\text{ sin } x \cos^4 x + 4 \sin^3 x + 2 \sin^3 x \cos^2 x\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3\text{ sin} x\left( 1 - \cos^4 x \right) + 2 \sin^3 x\left( 2 + \cos^2 x \right)\]

\[= 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3\text{ sin }x\left( 1 - \cos^2 x \right)\left( 1 + \cos^2 x \right) + 2 \sin^3 x\left( 2 + \cos^2 x \right)\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3 \sin^3 x\left( 1 + \cos^2 x \right) + 2 \sin^3 x\left( 2 + \cos^2 x \right)\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x\left[ 3\left( 1 + \cos^2 x \right) - 2\left( 2 + \cos^2 x \right) \right]\]
\[ = 5\text{  sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x\left[ 3 + 3 \cos^2 x - 4 - 2 \cos^2 x \right]\]
\[ = 5\text{ sin  }x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x \left[ \cos^2 x - 1 \right]\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x \times \left( - \sin^2 x \right)\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x + \sin^5 x\]
\[ = 5 \cos^4 x \text{ sin } x - 10 \cos^2 x \sin^3 x + \sin^5 x\]
\[ = RHS\]
\[\text{ Hence proved } .\]

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  Is there an error in this question or solution?
Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.2 [Page 36]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.2 | Q 8 | Page 36

RELATED QUESTIONS

Prove that:  \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]

 

Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]

 

Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]

 

Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]

 

Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x  \text{ cosec }  2 x\]

 

Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]


\[\tan 82\frac{1° }{2} = \left( \sqrt{3} + \sqrt{2} \right) \left( \sqrt{2} + 1 \right) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}\]

 


Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]

 

If  \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and }  \tan \frac{x}{2}\] . 

 

 


 If 0 ≤ x ≤ π and x lies in the IInd quadrant such that  \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and }  \tan\frac{x}{2}\]

 

 


If \[\tan A = \frac{1}{7}\]  and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4

 

 


Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]

 

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\] , prove that

(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]

 


If \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]

 

 


If \[a \cos2x + b \sin2x = c\]  has α and β as its roots, then prove that

(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\] 

 


Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]

 

\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{\pi}{3} - x \right) = 3 \cot 3x\]

 


Prove that: \[\cos 6° \cos 42°   \cos 66°    \cos 78° = \frac{1}{16}\]

 

Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]

 

If \[\cos 4x = 1 + k \sin^2 x \cos^2 x\] , then write the value of k.

 

If  \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .

 

 


The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\]  is 

  

If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then

 

The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is 

 

\[\frac{\sin 3x}{1 + 2 \cos 2x}\]   is equal to


The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is 

 

If  \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]

 


If \[\tan x = t\] then \[\tan 2x + \sec 2x =\]

 


If \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]

 

 


The greatest value of sin x cos x is ______.


The value of `cos  pi/5 cos  (2pi)/5 cos  (4pi)/5 cos  (8pi)/5`  is ______.


The value of `(sin 50^circ)/(sin 130^circ)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×