मराठी

Prove That: Sin X 2 Sin 7 X 2 + Sin 3 X 2 Sin 11 X 2 = Sin 2 X Sin 5 X .

Advertisements
Advertisements

प्रश्न

Prove that:
\[\sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2} = \sin 2x \sin 5x .\]

 

बेरीज
Advertisements

उत्तर

\[\text{ LHS }= \sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2}\]
\[ = \frac{1}{2}\left[ 2\sin\frac{x}{2}\sin\frac{7x}{2} + 2\sin\frac{3x}{2}\sin\frac{11x}{2} \right]\]
\[ = \frac{1}{2}\left[ \cos\left( \frac{7x}{2} - \frac{x}{2} \right) - \cos\left( \frac{7x}{2} + \frac{x}{2} \right) + \cos\left( \frac{11x}{2} - \frac{3x}{2} \right) - \cos\left( \frac{11x}{2} + \frac{3x}{2} \right) \right]\]
\[ = \frac{1}{2}\left[ \cos3x - \cos4x + \cos4x - \cos7x \right]\]
\[ = \frac{1}{2}\left[ \cos3x - \cos7x \right]\]
\[ = \frac{1}{2}\left[ - 2\sin\left( \frac{3x + 7x}{2} \right)\sin\left( \frac{3x - 7x}{2} \right) \right]\]
\[ = \frac{1}{2}\left[ - 2\sin\left( 5x \right)\sin\left( - 2x \right) \right]\]
\[ = \sin\left( 5x \right)\sin\left( 2x \right) =\text{ RHS }\] 
Hence, LHS = RHS

shaalaa.com
Transformation Formulae
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Transformation formulae - Exercise 8.2 [पृष्ठ १८]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.2 | Q 6.6 | पृष्ठ १८

संबंधित प्रश्‍न

Show that :

\[\sin 25^\circ \cos 115^\circ = \frac{1}{2}\left( \sin 140^\circ - 1 \right)\]

\[\text{ Prove that }4 \cos x \cos\left( \frac{\pi}{3} + x \right) \cos \left( \frac{\pi}{3} - x \right) = \cos 3x .\]

 


Prove that:
cos 20° cos 40° cos 80° = \[\frac{1}{8}\]

 


Express each of the following as the product of sines and cosines:
sin 12x + sin 4x


Prove that:
sin 105° + cos 105° = cos 45°


Prove that:
sin 40° + sin 20° = cos 10°


Prove that:

sin 51° + cos 81° = cos 21°

Prove that:

\[\cos\left( \frac{\pi}{4} + x \right) + \cos\left( \frac{\pi}{4} - x \right) = \sqrt{2} \cos x\]

 


Prove that:

\[\sin 65^\circ + \cos 65^\circ = \sqrt{2} \cos 20^\circ\]

Prove that:
cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A


Prove that:
 `sin A + sin 2A + sin 4A + sin 5A = 4 cos (A/2) cos((3A)/2)sin3A`


Prove that:

\[\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A} = \cot 8A\]

Prove that:

\[\frac{\sin A + \sin B}{\sin A - \sin B} = \tan \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right)\]

Prove that:

\[\frac{\cos A + \cos B}{\cos B - \cos A} = \cot \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right)\]

Prove that:

\[\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A\]

Prove that:

\[\frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A} = \tan 8A\]

Prove that:

\[\frac{\sin A + 2 \sin 3A + \sin 5A}{\sin 3A + 2 \sin 5A + \sin 7A} = \frac{\sin 3A}{\sin 5A}\]

Prove that:

\[\sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma) = 4 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\beta + \gamma}{2} \right) \sin \left( \frac{\gamma + \alpha}{2} \right)\]

 


\[\text{ If }\sin 2A = \lambda \sin 2B, \text{ prove that }\frac{\tan (A + B)}{\tan (A - B)} = \frac{\lambda + 1}{\lambda - 1}\]

 


\[\text{ If }\frac{\cos (A - B)}{\cos (A + B)} + \frac{\cos (C + D)}{\cos (C - D)} = 0, \text {Prove that }\tan A \tan B \tan C \tan D = - 1\]

 


If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ), prove that cot α cot β cot γ = cot δ

 

If y sin ϕ = x sin (2θ + ϕ), prove that (x + y) cot (θ + ϕ) = (y − x) cot θ.

 

If \[x \cos\theta = y \cos\left( \theta + \frac{2\pi}{3} \right) = z \cos\left( \theta + \frac{4\pi}{3} \right)\], prove that \[xy + yz + zx = 0\]

 

 


If sin 2A = λ sin 2B, then write the value of \[\frac{\lambda + 1}{\lambda - 1}\]


cos 35° + cos 85° + cos 155° =


sin 47° + sin 61° − sin 11° − sin 25° is equal to


Express the following as the sum or difference of sine or cosine:

`sin  "A"/8  sin  (3"A")/8`


Prove that:

(cos α – cos β)2 + (sin α – sin β)2 = 4 sin2 `((alpha - beta)/2)`


Prove that:

sin A sin(60° + A) sin(60° – A) = `1/4` sin 3A


Prove that:

2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0


Evaluate:

sin 50° – sin 70° + sin 10°


If tan θ = `1/sqrt5` and θ lies in the first quadrant then cos θ is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×