Question

prove it :sinΘsin(60-Θ)sin(60+Θ)=(sin3Θ)/4

Answer 1

❏Question :-

Prove that :

$\to \sin \theta \: \sin(60 - \theta) \: \sin(60 + \theta) = \frac{ \sin3 \theta}{4}$

❏Formula used :-

$\sf \star \boxed{ \bf \: {x}^{2} - {y}^{2} = (x + y)(x - y) }\\ \\ \star \boxed{ \bf\sin(x + y) = \sin x \cos y + \cos x \sin y} \\ \\ \star\boxed{ \bf\sin(x - y) = \sin x \cos y - \cos x \sin y} \: \\ \: \\ \star \boxed{ \sin3 \theta = 3 \sin \theta - 4 { \sin}^{3} \theta}$

❏Proof :-

Taking left hand side -

$\to \: \sin \theta \: \sin(60 - \theta) \: \sin(60 + \theta) \: \\ \\ \bf Apply \: the \: given \: formula \: \\ \\ \to \sin \theta \: ( \sin60 \cos \theta - \cos60 \sin \theta) \\ \: \: \: \: \times ( \sin60 \cos \theta - \cos60 \sin \theta) \\ \because \bf {x}^{2} - {y}^{2} = (x + y)(x - y) \\ \therefore \\ \to \sin \theta \bigg \{ {( \sin60 \cos \theta)}^{2} - {( \cos60 \sin \theta)}^{2} \bigg\} \\ \because \boxed{ \sin60 = \frac{ \sqrt{3} }{2} \: and \: \cos60 = \frac{1}{2} } \\ \therefore \\ \to \: \sin \theta \bigg \{ \frac{3}{4} { \cos}^{2} \theta - \frac{1}{4} { \sin}^{2} \theta \bigg \} \\ \\ \because \boxed{ { \cos}^{2} \theta = 1 - { \sin}^{2} \theta} \\ \therefore \\ \to \: \sin \theta \bigg \{ \frac{3}{4} (1 - { \sin}^{2} \theta) - \frac{1}{4} { \sin}^{2} \theta) \bigg \} \\ \\ \to \: \sin \theta \bigg \{ \frac{3}{4} - \frac{3}{4} { \sin}^{2} \theta - \frac{1}{4} { \sin}^{2} \theta \bigg \} \\ \\ \to \sin \theta \bigg \{ \frac{3}{4} - { \sin}^{2} \theta \bigg \} \\ \: \\ \to \: \sin \theta \bigg \{ \frac{3 - 4 { \sin}^{2} \theta }{4} \bigg \} \\ \\ \to \: \frac{3 \sin \theta - 4 { \sin}^{3} \theta }{4} \\ \\ \because \boxed{ \sin3 \theta = 3 \sin \theta - 4 \: { \sin}^{3} \theta} \\ \therefore \: \\ \to \: \frac{ \sin3 \theta}{4} \\ \\ \mathcal{ L.H.S = R.H.S \: } \\ \bf \: Hence \: Proved \:$

Answer 2

$\Large\underline\mathfrak{Question}$

$\sf \red{ prove} : \green{sinΘsin(60-Θ)sin(60+Θ)= \frac{1}{4} (sin3Θ)}$

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

$\red{1)} \blue{\boxed{\bf \: sin (A + B) sin (A - B) = sin^{2} A - sin^{2} B}}$

$\red{2)} \: \orange{\boxed{\bf \: sin60 \degree = \dfrac{ \sqrt{3} }{2}}}$

$\red{3)} \pink{\boxed{\bf3sin A - 4sin^{3} A=sin3A}}$

_____________________________

$\large\star{\underline{\tt{\red{Answer}}}}\star$

$\green{\rm \: solving\:L.H.S. : - -}$

$\red\leadsto\sf \: sinΘsin(60-Θ)sin(60+Θ) \\ \\ \red\leadsto\sf \: sinΘ[sin(60-Θ)sin(60+Θ)] \\ \\\red\leadsto\sf \: sinΘ[sin^{2}60 - sin^{2}Θ] \\ \\\red\leadsto\sf \:sinΘ[( \frac{ \sqrt{3}}{2})^{2} - sin^{2}Θ] \\ \\ \red\leadsto\sf sinΘ[ \frac{3}{4} - sin^{2}Θ] \\ \\ \red\leadsto\sf \: \frac{3sinΘ}{4} - sin^{3}Θ \\ \\ \red\leadsto\sf \: \frac{(3sinΘ - 4sin^{3}Θ)}{4} \\ \\\red\leadsto\sf \: \frac{sin3Θ}{4} \\ \\ \red\leadsto\sf \: \pink{\large\boxed{\boxed{\bold{\frac{1}{4}[sin3Θ]= \purple{RHS}}}}}$

$\huge\large\red{\boxed{\tt\blue{Hence} \: \orange,\green{Proved}}} \:$