Advertisements
Advertisements
प्रश्न
Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]
उत्तर
\[ = \frac{1}{2 \times 2\sin\frac{2\pi}{15}} \times \left( 2\sin\frac{4\pi}{15} \times \cos\frac{4\pi}{15} \right) \times \cos\frac{8\pi}{15} \times \cos\frac{16\pi}{15} \]
\[ = \frac{1}{2 \times 4\sin\frac{2\pi}{15}}\left( 2\sin\frac{8\pi}{15} \times \cos\frac{8\pi}{15} \right) \times \cos\frac{16\pi}{15}\]
\[ = \frac{1}{2 \times 8\sin\frac{2\pi}{15}}\left( 2\sin\frac{16\pi}{15} \times \cos\frac{16\pi}{15} \right)\]
\[ = \frac{1}{16\sin\frac{2\pi}{15}}\left( \sin\frac{32\pi}{15} \right)\]
\[= - \frac{1}{16\sin\frac{2\pi}{15}}\left( \sin2\pi - \frac{32\pi}{15} \right) \left[ \because \sin\left( 2\pi - \theta \right) = - sin\theta \right]\]
\[ = - \frac{1}{16\sin\frac{2\pi}{15}}\sin\left( - \frac{2\pi}{15} \right)\]
\[ = \frac{1}{16} = RHS\]
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
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 the IIIrd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2}, \sin 2x\] .
If \[\cos x = - \frac{3}{5}\] and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .
If \[\tan A = \frac{1}{7}\] and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B
Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\]
If \[\cos 4x = 1 + k \sin^2 x \cos^2 x\] , then write the value of k.
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
If \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
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
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
\[\frac{\sin 3x}{1 + 2 \cos 2x}\] 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 =
The value of \[\cos \left( 36° - A \right) \cos \left( 36° + A \right) + \cos \left( 54° - A \right) \cos \left( 54° + A \right)\] is
If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then
\[\cos2\alpha\] is equal to
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` is ______.
The value of cos12° + cos84° + cos156° + cos132° is ______.
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
The value of `(sin 50^circ)/(sin 130^circ)` is ______.
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
