हिंदी

The Value Of \[2 \Sin^2 B + 4 \Cos \Left( a + B \Right) \Sin a \Sin B + \Cos 2 \Left( a + B \Right)\] is

Advertisements
Advertisements

प्रश्न

The value of  \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is 

विकल्प

  • 0

  •  cos 3A

  • cos 2A

  •  none of these

MCQ
Advertisements

उत्तर

cos 2A

\[\text{ We have, }  \]

\[2 \sin^2 B + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B + \cos2\left( A + B \right)\]

\[ = 1 - \cos2B + \cos2\left( A + B \right) + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B\]

\[ = 1 + \left( \cos2\left( A + B \right) - \cos2B \right) + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B\]

\[ = 1 - 2\text{ sin } A\sin\left( A + 2B \right) + 4\cos\left( A + B \right) \text{ sin } A \text{ sin } B\]

\[ \left[ \because \text{ cos } C - \text{ cos } D = - 2\sin\frac{C + D}{2}\sin\frac{C - D}{2} \right]\]

\[ = 1 - 2\text{ sin } A\left[ \sin\left( A + 2B \right) - 2\text{ sin } B\cos\left( A + B \right) \right]\]

\[ = 1 - 2\text{ sin } A\left[ \sin\left( A + 2B \right) - \left\{ \sin\left( B + A + B \right) + \sin\left( B - \left( A + B \right) \right) \right\} \right] \]

\[ \left[ \because 2\text{ sin } C\text{ cos } D = \sin\left( C + D \right) + \sin\left( C - D \right) \right]\]

\[ = 1 - 2\text{ sin } A\left[ \sin\left( A + 2B \right) - \left\{ \sin\left( A + 2B \right) + \sin\left( - A \right) \right\} \right]\]

\[ = 1 - 2\text{ sin } A\left[ \text{ sin } A \right]\]

\[ = 1 - 2 \sin^2 A\]

\[ = \cos2A\]

 

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.5 [पृष्ठ ४४]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 22 | पृष्ठ ४४

संबंधित प्रश्न

Prove that:  \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]


Prove that:  \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]

 

Prove that: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]


Prove that:  \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]

 

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

 

Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]

 

\[\tan 82\frac{1° }{2} = \left( \sqrt{3} + \sqrt{2} \right) \left( \sqrt{2} + 1 \right) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}\]

 


If \[\text{ tan } x = \frac{b}{a}\] , then find the value of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\] . 

 

 


Prove that:  \[\cos 7°  \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]

 

Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]


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  \[\sin \alpha = \frac{4}{5} \text{ and }  \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]

 

If \[a \cos2x + b \sin2x = c\]  has α and β as its roots, then prove that 

(i) \[\tan\alpha + \tan\beta = \frac{2b}{a + c}\]

 


If \[a \cos2x + b \sin2x = c\]  has α and β as its roots, then prove that

(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\] 

 


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: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]

 

Prove that \[\left| \cos x \cos \left( \frac{\pi}{3} - x \right) \cos \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\]  for all values of x

 

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

  

Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]

  

Write the value of \[\cos^2 76°  + \cos^2 16°  - \cos 76° \cos 16°\] 

 

If  \[\text{ sin } x + \text{ cos } x = a\], then find the value of

\[\sin^6 x + \cos^6 x\] .
 

 


\[8 \sin\frac{x}{8} \cos \frac{x}{2}\cos\frac{x}{4} \cos\frac{x}{8}\]  is equal to 

 


The value of  \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is 

 

If  \[2 \tan \alpha = 3 \tan \beta, \text{ then }  \tan \left( \alpha - \beta \right) =\]

 


If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then

 

The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is 

 

If  \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\]  is equal to

 

\[2 \text{ cos } x - \ cos  3x - \cos 5x - 16 \cos^3 x \sin^2 x\]


The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]

 

If \[n = 1, 2, 3, . . . , \text{ then }  \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to

 


The value of `cos  pi/5 cos  (2pi)/5 cos  (4pi)/5 cos  (8pi)/5`  is ______.


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


The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.


If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos  theta/2` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×