If-p-sin-pi-18-sin-5pi-18-sin-7pi-18-find-the-value-of-p-

Question Number 105766 by bramlex last updated on 31/Jul/20

I f p = \sin (\frac{π}{18}) \sin (\frac{5 π}{18}) \sin (\frac{7 π}{18})

f i n d t h e v a l u e o f p .

Answered by som(math1967) last updated on 31/Jul/20

let \frac{π}{18} = α

∴ p = \sin α \sin 5 α \sin 7 α

= \sin α \sin (6 α - α) \sin (6 α + α)

= \sin α {\sin^{2} 6 α - \sin^{2} α}

= \sin \frac{π}{18} {\sin^{2} \frac{π}{3} - \sin^{2} \frac{π}{18}}

= \sin \frac{π}{18} {\frac{3}{4} - \sin^{2} \frac{π}{18}}

= \frac{1}{4} (3 \sin \frac{π}{18} - {4 \sin}^{3} \frac{π}{18})

= \frac{1}{4} \times \sin 3 . \frac{π}{18}

= \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

∴ p = \frac{1}{8} ans

Commented by 1549442205PVT last updated on 01/Aug/20

Third line ⇏ fourth line . Please, sir see

once again and check ?

Commented by bobhans last updated on 01/Aug/20

\sin α {\sin 6 α \cos α - \cos 6 α \sin α} {\sin

6 α \cos α + \cos 6 α \sin α}

\Leftrightarrow \sin α {{(\sin 6 α \cos α)}^{2} - {(\cos 6 α \sin α)}^{2}}

s o r r y s i r . y o u r a n s w e r s o m e t h i n g

w r o n g

Commented by som(math1967) last updated on 01/Aug/20

\sin (A - B) \sin (A + B)

= \sin^{2} A - \sin^{2} B

∴ \sin (6 α - α) \sin (6 α + α)

= \sin^{2} 6 α - \sin^{2} α

α = \frac{π}{18} \Rightarrow 6 α = \frac{π}{3}

∴ \sin α (\sin^{2} 6 α - \sin^{2} α)

= \sin α (\sin^{2} \frac{π}{3} - \sin^{2} α)

= \sin α (\frac{3}{4} - \sin^{2} α)

= \sin α (\frac{3 - {4 \sin}^{2} α}{4})

= \frac{3 \sin α - {4 \sin}^{3} α}{4}

= \frac{\sin 3 α}{4}

= \frac{\sin \frac{π}{6}}{4} [∵ 3 α = 3 \times \frac{π}{18} = \frac{π}{6}]

= \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

Commented by som(math1967) last updated on 01/Aug/20

\sin^{2} 6 α \cos^{2} α - \cos^{2} 6 α \sin^{2} α

= \sin^{2} 6 α (1 - \sin^{2} α)

- (1 - \sin^{2} 6 α) \sin^{2} α

= \sin^{2} 6 α - \sin^{2} 6 α \sin^{2} α

- \sin^{2} α + \sin^{2} 6 α \sin^{2} α

= \sin^{2} 6 α - \sin^{2} α

Commented by bobhans last updated on 01/Aug/20

\sin (A - B) \sin (A + B) = - \frac{1}{2} (- 2 \sin (A - B) \sin (A + B))

- \frac{1}{2} (\cos 2 A - \cos 2 B) = - \frac{1}{2} (1 - 2 \sin^{2} A - (1 - 2 \sin^{2} B))

= - \frac{1}{2} (- 2 \sin^{2} A + 2 \sin^{2} B)

= \sin^{2} A - \sin^{2} B .

h a h a h a . . y e s s i r . y o u a r e r i g h t

Commented by 1549442205PVT last updated on 03/Aug/20

Thank both sirs .

Commented by som(math1967) last updated on 01/Aug/20

Welcome