English

Prove That: Sin 4 X = 4 Sin X Cos 3 X − 4 Cos X Sin 3 X

Advertisements
Advertisements

Question

Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]

 
Numerical
Advertisements

Solution

\[LHS = sin 4x\]

\[ = 2\sin2x \cos2x \left( \because \sin2\theta = 2sin\theta cos\theta \right)\]

Now, using the identities

\[\sin2\alpha = 2\sin\alpha\cos\alpha \text{ and }  \cos2\alpha = \cos^2 \alpha - \sin^2 \alpha\], we get

\[LHS = 2(2\text{ sin } x \text{ cos } x) . ( \cos^2 x - \sin^2 x)\]

\[ = 4\text{ sin } x \cos^3 x - 4 \sin^3 x \text{ cos } x = RHS\]

\[\text{ Hence proved } .\]

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  Is there an error in this question or solution?
Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [Page 28]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 18 | Page 28

RELATED QUESTIONS

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


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

 

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


Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x  \text{ cosec }  2 x\]

 

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


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

 

 


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

 

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

(ii)  \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]

 


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 \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .

 

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


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

  

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

 

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 \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.

 
 

If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] . 

 

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

 

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

 
 

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

 


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

 


The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]

 

\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]


If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval


The value of  \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is 

  

\[\frac{\sin 3x}{1 + 2 \cos 2x}\]   is equal to


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


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

 

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

 


If \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]

 

 


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)`


Find the value of the expression `cos^4  pi/8 + cos^4  (3pi)/8 + cos^4  (5pi)/8 + cos^4  (7pi)/8`

[Hint: Simplify the expression to `2(cos^4  pi/8 + cos^4  (3pi)/8) = 2[(cos^2  pi/8 + cos^2  (3pi)/8)^2 - 2cos^2  pi/8 cos^2  (3pi)/8]`


The value of sin50° – sin70° + sin10° is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×