मराठी

Prove that Sin 3 X + Sin 2 X − Sin X = 4 Sin X Cos X 2 Cos 3 X 2

Advertisements
Advertisements

प्रश्न

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

संख्यात्मक
Advertisements

उत्तर

\[LHS = \sin3x + \sin2x - \text{ sin } x\]

\[ = \sin3x + 2\sin\left( \frac{2x - x}{2} \right)\cos\left( \frac{2x + x}{2} \right) \left[ \because \text{ sin } A - \text{ sin } B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]\]

\[ = \sin3x + 2\sin\left( \frac{x}{2} \right)\cos\left( \frac{3x}{2} \right)\]

\[ = 2\sin\left( \frac{3x}{2} \right)\cos\left( \frac{3x}{2} \right) + 2\sin\left( \frac{3x}{2} \right)\cos\frac{x}{2} \left[ \because \sin2A = 2\text{ sin } A\text{ cos } A \right]\]

\[ = 2\cos\left( \frac{3x}{2} \right)\left[ \sin\left( \frac{3x}{2} \right)\cos\left( \frac{x}{2} \right) \right]\]

\[ = 2\cos\left( \frac{3x}{2} \right)\left[ 2\sin\left( \frac{\frac{3x}{2} + \frac{x}{2}}{2} \right)\cos\left( \frac{\frac{3x}{2} - \frac{x}{2}}{2} \right) \right] \left[ \because \text{ sin } A + \text{ sin } B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]

\[ = 2\cos\left( \frac{3x}{2} \right)\left[ 2\sin\left( x \right)\cos\left( \frac{x}{2} \right) \right]\]

\[ = 4\sin\left( x \right)\cos\left( \frac{x}{2} \right)\cos\left( \frac{3x}{2} \right) = RHS\]

\[\text{ Hence proved } . \]

 

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [पृष्ठ २८]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 25 | पृष्ठ २८

संबंधित प्रश्‍न

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


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

 

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

 

Prove that:  \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]


Prove that:  \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]

 


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: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]

 

If \[\tan A = \frac{1}{7}\]  and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4

 

 


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 \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\] , prove that

(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]

 


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  \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]

 

If \[\cos \alpha + \cos \beta = \frac{1}{3}\]  and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]

 
 

 


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

 


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

 

\[\sin 5x = 5 \cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x\]

 


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:  \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\] 

 

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

 

Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]

 

If  \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .

 

 


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

 

\[\frac{\sec 8A - 1}{\sec 4A - 1} =\]

 


The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\]  is 

  

If \[\cos 2x + 2 \cos x = 1\]  then, \[\left( 2 - \cos^2 x \right) \sin^2 x\]  is equal to 

 
 

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

 

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

 

The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\]  is equal to

   

\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\]  is equal to


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

 


The greatest value of sin x cos x is ______.


If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = `(2b)/(a + c)`.

`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta =  (2tantheta)/(1 + tan^2theta)]`.


If θ lies in the first quadrant and cosθ = `8/17`, then find the value of cos(30° + θ) + cos(45° – θ) + cos(120° – θ).


If tanθ = `1/2` and tanΦ = `1/3`, then the value of θ + Φ is ______.


The value of `sin  pi/10  sin  (13pi)/10` is ______.

`["Hint: Use"  sin18^circ = (sqrt5 - 1)/4 "and"  cos36^circ = (sqrt5 + 1)/4]`


The value of `(sin 50^circ)/(sin 130^circ)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×