English

Prove That: Sin α + Sin β + Sin γ − Sin ( α + β + γ ) = 4 Sin ( α + β 2 ) Sin ( β + γ 2 ) Sin ( γ + α 2 )

Advertisements
Advertisements

Question

Prove that:

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

 

Sum
Advertisements

Solution

Consider LHS: 

\[ \sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma)\]

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

\[\]

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

\[\]

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

\[\]

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

\[\]

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

\[\]

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

\[\]

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

\[\]

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

\[\]

 = RHS

Hence, LHS = RHS.

shaalaa.com
Transformation Formulae
  Is there an error in this question or solution?
Chapter 8: Transformation formulae - Exercise 8.2 [Page 19]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 8 Transformation formulae
Exercise 8.2 | Q 9.1 | Page 19

RELATED QUESTIONS

Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]

 


Prove that tan 20° tan 30° tan 40° tan 80° = 1.


Express each of the following as the product of sines and cosines:
sin 5x − sin x


Express each of the following as the product of sines and cosines:
sin 2x + cos 4x


Prove that:
 cos 100° + cos 20° = cos 40°


Prove that:
sin 50° + sin 10° = cos 20°


Prove that:
 cos 55° + cos 65° + cos 175° = 0


Prove that:

sin 51° + cos 81° = cos 21°

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 3A + sin 2A − sin A = 4 sin A cos \[\frac{A}{2}\] \[\frac{3A}{2}\]

 


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

 


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:

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

Prove that:

\[\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A} = \cot 3A\]

 


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:
cos (A + B + C) + cos (A − B + C) + cos (A + B − C) + cos (− A + B + C) = 4 cos A cos Bcos C


\[\text{ If } \cos A + \cos B = \frac{1}{2}\text{ and }\sin A + \sin B = \frac{1}{4},\text{ prove that }\tan\left( \frac{A + B}{2} \right) = \frac{1}{2} .\]

 


If cosec A + sec A = cosec B + sec B, prove that tan A tan B = \[\cot\frac{A + B}{2}\].


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

 


If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), prove that tan A tan B tan C + tan D = 0.

 

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 A + sin B = α and cos A + cos B = β, then write the value of tan \[\left( \frac{A + B}{2} \right)\].

 

Write the value of the expression \[\frac{1 - 4 \sin 10^\circ \sin 70^\circ}{2 \sin 10^\circ}\]


If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.


sin 163° cos 347° + sin 73° sin 167° =


The value of sin 78° − sin 66° − sin 42° + sin 60° is ______.


If cos A = m cos B, then \[\cot\frac{A + B}{2} \cot\frac{B - A}{2}\]=

 

If A, B, C are in A.P., then \[\frac{\sin A - \sin C}{\cos C - \cos A}\]=

 

If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in


If \[\tan\alpha = \frac{x}{x + 1}\] and 

\[\tan\beta = \frac{1}{2x + 1}\], then
\[\tan\beta = \frac{1}{2x + 1}\] is equal to

 


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

cos 7θ sin 3θ


Express the following as the product of sine and cosine.

cos 2θ – cos θ


Prove that:

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


Prove that:

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


If secx cos5x + 1 = 0, where 0 < x ≤ `pi/2`, then find the value of x.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×