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: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
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 \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]
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)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
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 \[\frac{\pi}{2} < x < \pi,\] 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 \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
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 \[\sqrt{1 - \sin 2x}\] .
Write the value of \[\cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{4\pi}{7} .\]
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
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]
The value of \[\cos \left( 36° - A \right) \cos \left( 36° + A \right) + \cos \left( 54° - A \right) \cos \left( 54° + A \right)\] is
The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
The value of `(sin 50^circ)/(sin 130^circ)` is ______.
