हिंदी

sin 5 x sin x is equal to - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

विकल्प

  • \[16 \cos^4 x - 12 \cos^2 x + 1\]

     

  • \[16 \cos^4 x + 12 \cos^2 x + 1\]

     

  • \[16 \cos^4 x - 12 \cos^2 x - 1\]

     

  • \[16 \cos^4 x + 12 \cos^2 x - 1\]

     

MCQ
Advertisements

उत्तर

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

 

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.5 [पृष्ठ ४५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 34 | पृष्ठ ४५

संबंधित प्रश्न

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


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

 

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


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


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

 

Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]

 

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

 
 

 


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

 


If  \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .

 

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


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

 

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

 

Prove that:  \[\cos 78°  \cos 42°  \cos 36° = \frac{1}{8}\]


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

 

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

 

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

 

If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]

, then find the value of tan2A.

 

 


If \[\cos 2x + 2 \cos x = 1\]  then, \[\left( 2 - \cos^2 x \right) \sin^2 x\]  is equal to 

 
 

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

\[\cot A \cot B \cot C =\]
 

 


If  \[2 \tan \alpha = 3 \tan \beta, \text{ then }  \tan \left( \alpha - \beta \right) =\]

 


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

 

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

 


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

 

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


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


If \[n = 1, 2, 3, . . . , \text{ then }  \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to

 


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

 

 


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


Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A


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 cos248° – sin212° is ______.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×