English

sin 5 x sin x is equal to

Advertisements
Advertisements

Question

\[\frac{\sin 5x}{\sin x}\]  is equal to

 

Options

  • \[16 \cos^4 x - 12 \cos^2 x + 1\]

     

  • \[16 \cos^4 x + 12 \cos^2 x + 1\]

     

  • \[16 \cos^4 x - 12 \cos^2 x - 1\]

     

  • \[16 \cos^4 x + 12 \cos^2 x - 1\]

     

MCQ
Advertisements

Solution

\[\text{ To find } : \frac{\sin 5x}{\text{ sin } x}\]
\[\text{ Now} , \]
\[\sin5x = \sin\left( 3x + 2x \right)\]
\[ = \sin3x\cos2x + \cos3x\sin2x\]
\[ = \left( 3\text{ sin } x - 4 \sin^3 x \right)\left( 1 - 2 \sin^2 x \right) + \left( 4 \cos^3 x - 3\text{ cos } x \right)\left( 2\text{ sin } x\text{ cos } x \right)\]
\[ = \left( 3\sin x - 6 \sin^3 x - 4 \sin^3 x + 8 \sin^5 x \right) + 2\text{ sin } x \cos^2 x\left( 4 \cos^2 x - 3 \right)\]
\[ = \left( 3\sin x - 10 \sin^3 x + 8 \sin^5 x \right) + 2\text{ sin } x\left( 1 - \sin^2 x \right)\left[ 4\left( 1 - \sin^2 x \right) - 3 \right]\]
\[ = \left( 3\sin x - 10 \sin^3 x + 8 \sin^5 x \right) + \left( 2\text{ sin } x - 2 \sin^3 x \right)\left( 4 - 4 \sin^2 x - 3 \right)\]
\[ = \left( 3\sin x - 10 \sin^3 x + 8 \sin^5 x \right) + \left( 2\text{ sin } x - 8 \sin^3 x - 2 \sin^3 x + 8 \sin^5 x \right)\]
\[ = 5\text{ sin } x - 20 \sin^3 x + 16 \sin^5 x\]
\[ \therefore \frac{\sin 5x}{\text{ sin } x} = \frac{5\text{ sin } x - 20 \sin^3 x + 16 \sin^5 x}{\text{ sin } x} \]
\[ = 5 - 20 \sin^2 x + 16 \sin^4 x \]
\[ = 5 - 20\left( 1 - \cos^2 x \right) + 16 \left( 1 - \cos^2 x \right)^2 \]
\[ = 5 - 20 + 20 \cos^2 x + 16\left( 1 + \cos^4 x - 2 \cos^2 x \right)\]
\[ = 5 - 20 + 20 \cos^2 x + 16 + 16 \cos^4 x - 32 \cos^2 x\]
\[ = 16 \cos^4 x - 12 \cos^2 x + 1\]

 

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 9.5 [Page 45]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 34 | Page 45

RELATED QUESTIONS

Prove that:  \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]


Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]

 

Prove that:  \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]

 

Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]

 

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


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

 

Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]


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

 

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

 

 


Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]


If \[2 \tan \alpha = 3 \tan \beta,\]  prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .

 

If \[\cos \alpha + \cos \beta = \frac{1}{3}\]  and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]

 
 

 


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


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

  

Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]

  

Prove that:  \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\] 

 

Prove that: \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]

 

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

 

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 in a  \[∆ ABC, \tan A + \tan B + \tan C = 0\], then

\[\cot A \cot B \cot C =\]
 

 


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

 

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


If  \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\]  is equal to

 

\[2 \text{ cos } x - \ cos  3x - \cos 5x - 16 \cos^3 x \sin^2 x\]


\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\]  is equal to


The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is

 

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

 

 


If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.


The value of sin 20° sin 40° sin 60° sin 80° is ______.


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


If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = `(2b)/(a + c)`.

`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta =  (2tantheta)/(1 + tan^2theta)]`.


If θ lies in the first quadrant and cosθ = `8/17`, then find the value of cos(30° + θ) + cos(45° – θ) + cos(120° – θ).


Find the value of the expression `cos^4  pi/8 + cos^4  (3pi)/8 + cos^4  (5pi)/8 + cos^4  (7pi)/8`

[Hint: Simplify the expression to `2(cos^4  pi/8 + cos^4  (3pi)/8) = 2[(cos^2  pi/8 + cos^2  (3pi)/8)^2 - 2cos^2  pi/8 cos^2  (3pi)/8]`


The value of cos248° – sin212° is ______.

[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×