Advertisements
Advertisements
Question
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
Advertisements
Solution
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is `underlinebb(1/8)`.
Explanation:
Given that: k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`
⇒ k = sin10°. sin50°. sin70°
⇒ k = sin10° sin(90° – 40°) sin(90° – 20°)
⇒ k = sin10° cos40° cos20°
⇒ k = `sin10^circ . 1/2 [2 cos 40^circ cos 20^circ]`
⇒ k = `sin 10^circ . 1/2 [cos(40^circ + 20^circ) + cos(40^circ - 20^circ)]`
⇒ k = `1/2 sin 10^circ [cos 60^circ + cos 20^circ]`
⇒ k = `1/2 sin 10^circ(1/2 + cos 20^circ)`
⇒ k = `1/4 sin 10^circ + 1/2 sin 10^circ . cos 20^circ`
⇒ k = `1/4 sin 10^circ + 1/4(2 sin 10^circ cos 20^circ)`
⇒ k = `1/4 sin 10^circ + 1/4[sin(10^circ + 20^circ) + sin(10^circ - 20^circ)]`
⇒ k = `1/4 sin 10^circ + 1/4[sin30^circ + sin(-10^circ)]`
⇒ k = `1/4 sin 10^circ + 1/4 sin 30^circ - 1/4 sin 10^circ`
= `1/4 sin 30^circ`
= `1/4 xx 1/2`
= `1/8`
APPEARS IN
RELATED QUESTIONS
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
If \[\cos x = - \frac{3}{5}\] and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .
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 \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(i) \[\tan\alpha + \tan\beta = \frac{2b}{a + c}\]
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: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
If \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .
If \[\cos 2x + 2 \cos x = 1\] then, \[\left( 2 - \cos^2 x \right) \sin^2 x\] is equal to
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
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
The value of sin 20° sin 40° sin 60° sin 80° is ______.
Find the value of the expression `cos^4 pi/8 + cos^4 (3pi)/8 + cos^4 (5pi)/8 + cos^4 (7pi)/8`
[Hint: Simplify the expression to `2(cos^4 pi/8 + cos^4 (3pi)/8) = 2[(cos^2 pi/8 + cos^2 (3pi)/8)^2 - 2cos^2 pi/8 cos^2 (3pi)/8]`
If tanθ = `1/2` and tanΦ = `1/3`, then the value of θ + Φ 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 sin50° – sin70° + sin10° is equal to ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
