English

Sin 3 X + Sin 3 ( 2 π 3 + X ) + Sin 3 ( 4 π 3 + X ) = − 3 4 Sin 3 X - Mathematics

Advertisements
Advertisements

Question

\[\sin^3 x + \sin^3 \left( \frac{2\pi}{3} + x \right) + \sin^3 \left( \frac{4\pi}{3} + x \right) = - \frac{3}{4} \sin 3x\]

 

Numerical
Advertisements

Solution

\[LHS = \sin^3 x + \sin^3 \left( \frac{2\pi}{3} + x \right) + \sin^3 \left( \frac{4\pi}{3} + x \right)\]
\[ = \frac{3\text{ sin } x - \sin3x}{4} + \frac{3\sin\left( \frac{2\pi}{3} + x \right) - \sin3\left( \frac{2\pi}{3} + x \right)}{4} + \frac{3\sin\left( \frac{4\pi}{3} + x \right) - \sin3\left( \frac{4\pi}{3} + x \right)}{4} \]
\[ \left[ \sin^3 \theta = \frac{3sin\theta - \sin3\theta}{4} \right]\]
\[ = \frac{3\text{ sin } x - \sin3x}{4} + \frac{3\sin\left\{ \pi - \left( \frac{2\pi}{3} + x \right) \right\} - \sin\left( 2\pi + 3x \right)}{4} + \frac{3\sin\left\{ \pi + \left( \frac{\pi}{3} + x \right) \right\} - \sin\left( 4\pi + 3x \right)}{4}\]
\[ = \frac{1}{4}\left[ \left( 3\text{ sin } x - \sin3x \right) + \left\{ 3\sin\left( \frac{\pi}{3} - x \right) - \sin3x \right\} - \left\{ 3\sin\left( \frac{\pi}{3} + x \right) + \sin3x \right\} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - \sin3x + 3\sin\left( \frac{\pi}{3} - x \right) - 3\sin\left( \frac{\pi}{3} + x \right) - \sin3x - \sin3x \right]\]

\[= \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 3\left\{ \sin\left( \frac{\pi}{3} - x \right) - \sin\left( \frac{\pi}{3} + x \right) \right\} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 3\left\{ 2\cos\frac{\frac{\pi}{3} - x + \frac{\pi}{3} + x}{2}\sin\frac{\frac{\pi}{3} - x - \frac{\pi}{3} - x}{2} \right\} \right]\]
\[ \left[ \because sinC - sinD = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 6\cos\frac{\pi}{3}\sin\left( - x \right) \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\text{ sin } 3x - 3\text{ sin } x \right]\]
\[ = - \frac{3}{4}\text{ sin } x\]
\[ = RHS\]
\[\text{ Hence proved } .\]

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 37]

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 9 | Page 37

RELATED QUESTIONS

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


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

 

Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]


Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]


Prove that:  \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]

 


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

 

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

 

Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]

 

If  \[\sin \alpha = \frac{4}{5} \text{ and }  \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]

 

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

(i) \[\tan\alpha + \tan\beta = \frac{2b}{a + c}\]

 


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 `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`


\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\] 


Prove that \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\]  for all values of x

 
 

Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]

  

Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]

 

Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78°  = \frac{1}{16}\]

 

If \[\tan\frac{x}{2} = \frac{m}{n}\] , then write the value of m sin x + n cos x.

 

 


If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] . 

 

In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.

 

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

 

 


\[8 \sin\frac{x}{8} \cos \frac{x}{2}\cos\frac{x}{4} \cos\frac{x}{8}\]  is equal to 

 


For all real values of x, \[\cot x - 2 \cot 2x\] is equal to 

 

The value of  \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is 

 

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha - \cos \beta = b \text{ then }  \tan \frac{\alpha - \beta}{2} =\]

 


\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]


The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\]  is equal to

   

If  \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]


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


If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ 
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]


The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.


The value of `sin  pi/10  sin  (13pi)/10` is ______.

`["Hint: Use"  sin18^circ = (sqrt5 - 1)/4 "and"  cos36^circ = (sqrt5 + 1)/4]`


If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos  theta/2` is ______.


If A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×