Advertisements
Advertisements
प्रश्न
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
Advertisements
उत्तर
\[LHS = \cos36° \cos42° \cos60° \cos78° \]
\[ = \frac{1}{2}\cos36° \cos60° \left( 2\cos42° \cos78° \right)\]
\[ \left[ 2\text{ cos } A\text{ cos } B = \cos\left( A + B \right) + \cos\left( A - B \right) \right]\]
\[ = \frac{1}{2}\left( \frac{\sqrt{5} + 1}{4} \right) \times \frac{1}{2}\left( \cos120° + \cos36° \right) \]
\[ = \left( \frac{\sqrt{5} + 1}{16} \right)\left( - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{\left( \sqrt{5} + 1 \right) \left( \sqrt{5} - 1 \right)}{64}\]
\[ = \frac{5 - 1}{64}\]
\[ = \frac{1}{16}\]
\[ = RHS\]
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x \text{ cosec } 2 x\]
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
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 \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , 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
(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: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]
Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
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 \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
If \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
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 \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is
If \[n = 1, 2, 3, . . . , \text{ then } \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to
The value of `cos^2 48^@ - sin^2 12^@` is ______.
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
The greatest value of sin x cos x is ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
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).
The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.
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 cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
