English

Prove that: cos π 65 cos 2 π 65 cos 4 π 65 cos 8 π 65 cos 16 π 65 cos 32 π 65 = 1 64

Advertisements
Advertisements

Question

Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]

 
Numerical
Advertisements

Solution

\[LHS = \cos\frac{\pi}{65} \cos\frac{2\pi}{65}\cos\frac{4\pi}{65} \cos\frac{8\pi}{65}\cos\frac{16\pi}{65} \cos\frac{32\pi}{65}\]

On dividing and multiplying by  \[2\sin\frac{\pi}{65}\] , we get

\[= \frac{1}{2\sin\frac{\pi}{65}} \times 2\sin\frac{\pi}{65} \times \cos\frac{\pi}{65} \times \cos\frac{2\pi}{65} \times \cos\frac{4\pi}{65} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]
\[ = \frac{2 \times \sin2\frac{\pi}{65}}{2 \times 2\sin\frac{\pi}{65}} \times \cos\frac{2\pi}{65} \times \cos\frac{4\pi}{65} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65} \]
\[ = \frac{2 \times \sin4\frac{\pi}{65}}{2 \times 4\sin\frac{\pi}{65}} \times \cos\frac{4\pi}{65} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]
\[ = \frac{2 \times \sin8\frac{\pi}{65}}{2 \times 8\sin\frac{\pi}{65}} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]

\[= \frac{2 \times \sin16\frac{\pi}{65}}{2 \times 16\sin\frac{\pi}{65}} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]
\[ = \frac{2 \times \sin32\frac{\pi}{65}}{2 \times 32\sin\frac{\pi}{65}} \times \cos\frac{32\pi}{65}\]
\[ = \frac{\sin64\frac{\pi}{65}}{64\sin\frac{\pi}{65}} = \frac{\sin\left( \pi - \frac{\pi}{65} \right)}{64\sin\frac{\pi}{65}}\]
\[ = \frac{\sin\frac{\pi}{65}}{64\sin\frac{\pi}{65}} \left[ \because \sin\left( \pi - \theta \right) = sin\theta \right]\]
\[ = \frac{1}{64} = RHS\]
\[\text{ Hence proved } .\]

 
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 29]

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 36 | Page 29

RELATED QUESTIONS

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


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

 


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

 

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

 

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


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


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

 

 If 0 ≤ x ≤ π and x lies in the IInd quadrant such that  \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and }  \tan\frac{x}{2}\]

 

 


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

, find the value of sin 4x.

 

 


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

 

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 \[a \cos2x + b \sin2x = c\]  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 \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .

 

Prove that:  \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]

 

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

 

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

 


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

 


Prove that:  \[\cos 78°  \cos 42°  \cos 36° = \frac{1}{8}\]


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

 

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

 

If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .

 

 


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

\[\sin^6 x + \cos^6 x\] .
 

 


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 

 
 

If  \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]


If  \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]

 


The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is 

 

If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ 
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]


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 `sin  pi/10  sin  (13pi)/10` is ______.

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


If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×