Advertisements
Advertisements
Question
Prove that: \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]
Advertisements
Solution
\[LHS = \sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}\]
\[ = \frac{1}{2}\left( 2 \sin\frac{\pi}{5} \sin\frac{4\pi}{5} \right)\frac{1}{2}\left( 2 \sin\frac{2\pi}{5} \sin\frac{3\pi}{5} \right)\]
\[ = \frac{1}{4}\left( \cos\left( \frac{\pi}{5} - \frac{4\pi}{5} \right) - \cos\left( \frac{\pi}{5} + \frac{4\pi}{5} \right) \right)\left( \cos\left( \frac{2\pi}{5} - \frac{3\pi}{5} \right) - \cos\left( \frac{2\pi}{5} + \frac{3\pi}{5} \right) \right)\]
\[ = \frac{1}{4}\left( \cos\left( \frac{- 3\pi}{5} \right) - \cos\left( \frac{5\pi}{5} \right) \right)\left( \cos\left( \frac{- \pi}{5} \right) - \cos\left( \frac{5\pi}{5} \right) \right)\]
\[ = \frac{1}{4}\left( \cos\left( \frac{3\pi}{5} \right) - \cos\left( \pi \right) \right)\left( \cos\left( \frac{\pi}{5} \right) - \cos\left( \pi \right) \right)\]
\[ = \frac{1}{4}\left( \cos\left( \frac{3\pi}{5} \right) + 1 \right)\left( \cos\left( \frac{\pi}{5} \right) + 1 \right)\]
\[ = \frac{1}{4}\left( \cos\left( \pi - \frac{2\pi}{5} \right) + 1 \right)\left( \cos\left( \frac{\pi}{5} \right) + 1 \right)\]
\[ = \frac{1}{4}\left( - \cos\left( \frac{2\pi}{5} \right) + 1 \right)\left( \left( \frac{\sqrt{5} + 1}{4} \right) + 1 \right) \left( \because \cos \frac{\pi}{5} = \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{1}{4}\left( - \left( \frac{\sqrt{5} - 1}{4} \right) + 1 \right)\left( \left( \frac{\sqrt{5} + 1}{4} \right) + 1 \right) \left( \because \cos \frac{2\pi}{5} = \frac{\sqrt{5} - 1}{4} \right)\]
\[ = \frac{1}{4}\left( - \left( \frac{\sqrt{5} - 1}{4} \right)\left( \frac{\sqrt{5} + 1}{4} \right) - \left( \frac{\sqrt{5} - 1}{4} \right) + \left( \frac{\sqrt{5} + 1}{4} \right) + 1 \right)\]
\[ = \frac{1}{4}\left( - \left( \frac{\left( \sqrt{5} \right)^2 - 1}{16} \right) + \left( \frac{\sqrt{5} + 1 - \sqrt{5} + 1}{4} \right) + 1 \right)\]
\[ = \frac{1}{4}\left( - \left( \frac{4}{16} \right) + \left( \frac{2}{4} \right) + 1 \right)\]
\[ = \frac{1}{4}\left( - \frac{1}{4} + \frac{2}{4} + 1 \right)\]
\[ = \frac{1}{4}\left( \frac{- 1 + 2 + 4}{4} \right)\]
\[ = \frac{5}{16}\]
\[ = RHS\]
Thus, LHS = RHS
Hence,
APPEARS IN
RELATED QUESTIONS
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
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: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]
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 \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan \frac{x}{2}\] .
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]
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 \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
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 \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
Write the value of \[\cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{4\pi}{7} .\]
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\] and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is
\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]
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
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
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 ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`
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 sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
