English

Prove That: Sin 5 π 18 − Cos 4 π 9 = √ 3 Sin π 9

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

\[LHS = \sin\left( \frac{5\pi}{18} \right) - \cos\frac{4\pi}{9}\]
\[ = \sin\left( \frac{5\pi}{18} \right) - \cos\left( \frac{\pi}{2} - \frac{\pi}{18} \right)\]
\[ = \sin\left( \frac{5\pi}{18} \right) - \sin\left( \frac{\pi}{18} \right)\]
\[ = 2\sin\left( \frac{\frac{5\pi}{18} - \frac{\pi}{18}}{2} \right)\cos\left( \frac{\frac{5\pi}{18} + \frac{\pi}{18}}{2} \right) \left[ \because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]\]
\[ = 2\sin\left( \frac{\pi}{9} \right)\cos\frac{\pi}{6}\]
\[ = 2\sin\left( \frac{\pi}{9} \right)\cos\frac{\pi}{6}\]
\[ = 2 \times \frac{\sqrt{3}}{2}\sin\left( \frac{\pi}{9} \right)\]
\[ = \sqrt{3}\sin\left( \frac{\pi}{9} \right) = RHS\]
Hence, LHS = RHS.

shaalaa.com
Transformation Formulae
  Is there an error in this question or solution?
Chapter 8: Transformation formulae - Exercise 8.2 [Page 17]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 8 Transformation formulae
Exercise 8.2 | Q 3.5 | Page 17

RELATED QUESTIONS

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

 


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


Express each of the following as the product of sines and cosines:
 cos 12x + cos 8x


Express each of the following as the product of sines and cosines:
 cos 12x - cos 4x


Prove that:
sin 38° + sin 22° = sin 82°


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


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


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


Prove that:

sin 80° − cos 70° = cos 50°

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

 


Prove that:

cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −\[\frac{3}{4}\]

 


Prove that \[\cos x \cos \frac{x}{2} - \cos 3x \cos\frac{9x}{2} = \sin 7x \sin 8x\]

Prove that:

\[\frac{\sin A - \sin B}{\cos A + \cos B} = \tan\frac{A - B}{2}\]

Prove that:

\[\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A} = \cot 6A\]

Prove that:

\[\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A} = \tan A\]

Prove that:

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

Prove that:

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

 


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


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


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 \[\left( \frac{A + B}{2} \right)\].

 

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


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

 

 


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


cos 35° + cos 85° + cos 155° =


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


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

 

Express the following as the product of sine and cosine.

sin A + sin 2A


Prove that:

tan 20° tan 40° tan 80° = `sqrt3`.


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


Evaluate-

cos 20° + cos 100° + cos 140°


Evaluate:

sin 50° – sin 70° + sin 10°


If cosec A + sec A = cosec B + sec B prove that cot`(("A + B"))/2` = tan A tan B.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×