मराठी

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

Advertisements
Advertisements

प्रश्न

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

 

संख्यात्मक
Advertisements

उत्तर

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

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

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.2 [पृष्ठ ३६]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.2 | Q 8 | पृष्ठ ३६

संबंधित प्रश्‍न

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


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

 

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

 


 If \[\sin x = \frac{4}{5}\] and \[0 < x < \frac{\pi}{2}\]

, find the value of sin 4x.

 

 


If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]

 

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

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

 


Prove that:  \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]

 

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

 


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


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

 

 


If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.

 
 

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

 

Write the value of \[\cos^2 76°  + \cos^2 16°  - \cos 76° \cos 16°\] 

 

If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .

 

 


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

 
 

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

 

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

 


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

   

If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =

 

The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is 


The value of \[\cos \left( 36°  - A \right) \cos \left( 36° + A \right) + \cos \left( 54°  - A \right) \cos \left( 54°  + A \right)\] is 

 

The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is 

 

The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]

 

If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then

\[\cos2\alpha\]   is equal to

 

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


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 cos12° + cos84° + cos156° + cos132° is ______.


The value of sin50° – sin70° + sin10° is equal to ______.


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×