मराठी

Prove That: Cos 55° + Cos 65° + Cos 175° = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Consider LHS: 
\[\cos 55^\circ + \cos 65^\circ + \cos 175^\circ\]
\[ = 2\cos \left( \frac{55^\circ + 65^\circ}{2} \right) \cos \left( \frac{55^\circ - 65^\circ}{2} \right) + \cos 175^\circ \left\{ \because \cos A + \cos B = 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\cos 60^\circ \cos\left( - 5^\circ \right) + \cos 175^\circ\]
\[ = 2 \times \frac{1}{2}\cos 5^\circ + \cos 175^\circ\]
\[ = \cos 5^\circ + \cos 175^\circ\]
\[ = 2\cos \left( \frac{5^\circ + 175^\circ}{2} \right) \cos \left( \frac{5^\circ - 175^\circ}{2} \right)\]
\[ = 2\cos 90^\circ \cos 85^\circ\]
\[ = 0\]
Hence, LHS = RHS.

shaalaa.com
Transformation Formulae
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Transformation formulae - Exercise 8.2 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.2 | Q 3.1 | पृष्ठ १७

संबंधित प्रश्‍न

Prove that:
 sin 20° sin 40° sin 60° sin 80° = \[\frac{3}{16}\]

 


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


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


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


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


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


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

Prove that:

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

 


Prove that:

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

Prove that:

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

Prove that:

\[\frac{\sin \left( \theta + \phi \right) - 2 \sin \theta + \sin \left( \theta - \phi \right)}{\cos \left( \theta + \phi \right) - 2 \cos \theta + \cos \left( \theta - \phi \right)} = \tan \theta\]

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

 


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

 


Prove that:
 sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0


If (cos α + cos β)2 + (sin α + sin β)2 = \[\lambda \cos^2 \left( \frac{\alpha - \beta}{2} \right)\], write the value of λ. 


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 A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].

 

 


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


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) =

 

 


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


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


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

 

Express the following as the product of sine and cosine.

sin 6θ – sin 2θ


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


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


Evaluate-

cos 20° + cos 100° + cos 140°


If cos A + cos B = `1/2` and sin A + sin B = `1/4`, prove that tan `(("A + B")/2) = 1/2`


Find the value of tan22°30′. `["Hint:"  "Let" θ = 45°, "use" tan  theta/2 = (sin  theta/2)/(cos  theta/2) = (2sin  theta/2 cos  theta/2)/(2cos^2  theta/2) = sintheta/(1 + costheta)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×