मराठी

The value of cos248° – sin212° is ______. [Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)] - Mathematics

Advertisements
Advertisements

प्रश्न

The value of cos248° – sin212° is ______.

[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]

पर्याय

  • `(sqrt(5) + 1)/8`

  • `(sqrt(5) - 1)/8`

  • `(sqrt(5) + 1)/5`

  • `(sqrt(5) + 1)/(2sqrt(2)`

MCQ
रिकाम्या जागा भरा

उत्तर

The value of cos248° – sin212° is `bbunderline((sqrt(5) + 1)/8)`.

Explanation:

The given expression is cos248° – sin212°.

cos248° – sin212° = cos(48° + 12°).cos(48° – 12°)   ......[∵ cos2A – sin2B = cos(A + B).cos(A – B)]

= cos 60°.cos 36°

= `1/2 xx (sqrt(5) + 1)/4`

= `(sqrt(5) + 1)/8`

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Exercise [पृष्ठ ५८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Exercise | Q 56 | पृष्ठ ५८

संबंधित प्रश्‍न

Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]

 

Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]

 

Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]


Show that: \[3 \left( \sin x - \cos x \right)^4 + 6 \left( \sin x + \cos \right)^2 + 4 \left( \sin^6 x + \cos^6 x \right) = 13\]


Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]


Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]


 If  \[\cos x = - \frac{3}{5}\]  and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .

 

 


 If \[\sin x = \frac{4}{5}\] and \[0 < x < \frac{\pi}{2}\]

, find the value of sin 4x.

 

 


Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]

 

If \[2 \tan \alpha = 3 \tan \beta,\]  prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .

 

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\] , prove that 
(i)\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]


If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]

 

If  \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .

 

Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`


\[\sin^3 x + \sin^3 \left( \frac{2\pi}{3} + x \right) + \sin^3 \left( \frac{4\pi}{3} + x \right) = - \frac{3}{4} \sin 3x\]

 


Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]

  

Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]

 

Prove that: \[\cos 6° \cos 42°   \cos 66°    \cos 78° = \frac{1}{16}\]

 

Prove that : \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]

 

Write the value of \[\cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{4\pi}{7} .\]

  

If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\]  and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]

 

 


If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]


If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =

 

The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is 


The value of \[\cos \left( 36°  - A \right) \cos \left( 36° + A \right) + \cos \left( 54°  - A \right) \cos \left( 54°  + A \right)\] is 

 

\[\frac{\sin 5x}{\sin x}\]  is equal to

 


If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then

\[\cos2\alpha\]   is equal to

 

If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×