हिंदी

Cos 40° + Cos 80° + Cos 160° + Cos 240° = - Mathematics

Advertisements
Advertisements

प्रश्न

cos 40° + cos 80° + cos 160° + cos 240° =

विकल्प

  • 0

  • 1

  • \[\frac{1}{2}\]

     

  • \[- \frac{1}{2}\]

     

MCQ
योग
Advertisements

उत्तर

\[- \frac{1}{2}\]

\[\cos40^\circ + \cos80^\circ + \cos160^\circ + \cos240^\circ\]
\[ = 2\cos\left( \frac{40^\circ + 80^\circ}{2} \right)\cos\left( \frac{40^\circ - 80^\circ}{2} \right) + \cos160^\circ - \cos\left( 180^\circ + 60^\circ \right) \left[ \because \cos A + \cos B = 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]
\[ = 2\cos60^\circ \cos\left( - 20^\circ \right) + \cos160^\circ - \frac{1}{2}\]
\[ = 2 \times \frac{1}{2}\cos20^\circ + \cos160^\circ - \frac{1}{2}\]
\[ = - \cos\left( 180 - 20 \right)^\circ + \cos160^\circ - \frac{1}{2}\]
\[ = - \frac{1}{2}\]

shaalaa.com
Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.4 [पृष्ठ २१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.4 | Q 1 | पृष्ठ २१

संबंधित प्रश्न

Prove that:

\[2\cos\frac{5\pi}{12}\cos\frac{\pi}{12} = \frac{1}{2}\]

Prove that: 

\[2\sin\frac{5\pi}{12}\cos\frac{\pi}{12} = \frac{\sqrt{3} + 2}{2}\]

Show that :

\[\sin 50^\circ \cos 85^\circ = \frac{1 - \sqrt{2} \sin 35^\circ}{2\sqrt{2}}\]

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

 


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


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 55° + cos 65° + cos 175° = 0


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


Prove that:

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

Prove that:

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

 


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:

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

Prove that:

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

 


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:

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

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


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

 


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

 

If (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 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 (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 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:

cos(60° + A) sin(120° + A)


Prove that:

tan 20° tan 40° tan 80° = `sqrt3`.


Prove that:

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


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 2"A" - cos 3"A")/(sin "2A" + sin "3A") = tan  "A"/2`


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


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×