हिंदी

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

प्रश्न

Prove that:

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

 

योग
Advertisements

उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.2 [पृष्ठ १९]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 9.1 | पृष्ठ १९

संबंधित प्रश्न

Prove that:
cos 10° cos 30° cos 50° cos 70° = \[\frac{3}{16}\]


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

 


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


Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0


Prove that:
 sin 23° + sin 37° = cos 7°


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


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


Prove that:
cos 20° + cos 100° + cos 140° = 0


Prove that:
\[\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}\]


Prove that:

\[\cos\frac{\pi}{12} - \sin\frac{\pi}{12} = \frac{1}{\sqrt{2}}\]

 


Prove that: 
cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A


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


Prove that:

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

 


Prove that:

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

Prove that:

\[\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A} = \tan 2A\]

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 = \[\cot\frac{A + B}{2}\].


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

 

 


Write the value of sin \[\frac{\pi}{12}\] sin \[\frac{5\pi}{12}\].


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

 

If cos A = m cos B, then write the value of \[\cot\frac{A + B}{2} \cot\frac{A - B}{2}\].

 

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


If sin 2 θ + sin 2 ϕ = \[\frac{1}{2}\] and cos 2 θ + cos 2 ϕ = \[\frac{3}{2}\], then cos2 (θ − ϕ) =

 

 


The value of sin 50° − sin 70° + sin 10° is equal to


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

 

If sin x + sin y = \[\sqrt{3}\] (cos y − cos x), then sin 3x + sin 3y =

 


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

`sin  "A"/8  sin  (3"A")/8`


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

cos(60° + A) sin(120° + A)


Prove that:

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


Prove that:

sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0


Prove that:

`(cos 7"A" +cos 5"A")/(sin 7"A" −sin 5"A")` = cot A


Prove that cos 20° cos 40° cos 60° cos 80° = `3/16`.


Evaluate-

cos 20° + cos 100° + cos 140°


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×