Advertisements
Advertisements
प्रश्न
If \[n = 1, 2, 3, . . . , \text{ then } \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to
पर्याय
\[\frac{\sin 2n \alpha}{2n \sin \alpha}\]
- \[\frac{\sin 2^n \alpha}{2^n \sin 2^{n - 1} \alpha}\]
\[\frac{\sin 4^{n - 1} \alpha}{4^{n - 1} \sin \alpha}\]
- \[\frac{\sin 2^n \alpha}{2^n \sin \alpha}\]
Advertisements
उत्तर
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]
Prove that: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
Prove that: \[\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} + \cos^2 \frac{5\pi}{8} + \cos^2 \frac{7\pi}{8} = 2\]
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: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
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 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 \[\tan A = \frac{1}{7}\] and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B
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 \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]
If \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
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
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 in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
\[\frac{\sin 3x}{1 + 2 \cos 2x}\] is equal to
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` 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)]
If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`
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 ______.
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
