Question

prove that sin 10.sin 50.sin70=root of 3/8

Answer 1

Appropriate Question :-

$\rm :\longmapsto\:\sf \: Prove \: that \: sin10\degree \: sin50\degree \: sin70\degree = \dfrac{1}{8}$

$\large\underline{\sf{Solution-}}$

We know,

$\boxed{ \sf \: sinx \: sin(60 \degree - x) \: sin(60\degree + x) = \dfrac{1}{4}sin3x}$

Let first derive this result.

Consider,

$\rm :\longmapsto\:sinx \: sin(60\degree - x)sin(60\degree + x)$

$\: \: \rm = \: \: sinx( {sin}^{2}60\degree - {sin}^{2}x)$

$\: \: \: \: \: \boxed{ \sf \: \because \: sin(x + y)sin(x - y) = {sin}^{2}x - {sin}^{2}y}$

$\: \: \rm = \: \: sinx \: \bigg( {\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }^{2} - {sin}^{2}x\bigg)$

$\: \: \rm = \: \: sinx \: \bigg(\dfrac{3}{4} - {sin}^{2}x\bigg)$

$\: \: \rm = \: \: sinx \: \bigg(\dfrac{3 - 4 {sin}^{2}x}{4}\bigg)$

$\: \: \rm = \: \: \: \bigg(\dfrac{3sinx - 4 {sin}^{3}x}{4}\bigg)$

$\: \: \rm = \: \: \dfrac{1}{4}sin3x$

Hence,

$\rm :\longmapsto\:sinx \: sin(60\degree - x)sin(60\degree + x) = \dfrac{1}{4}sin3x$

Put x = 10° in the above result, we get

$\rm :\longmapsto\:sin10\degree \: sin(60\degree - 10\degree)sin(60\degree + 10\degree) = \dfrac{1}{4}sin30\degree$

$\rm :\longmapsto\:sin10\degree \: sin50\degree \: sin70\degree = \dfrac{1}{4} \times \dfrac{1}{2}$

$\rm :\longmapsto\:sin10\degree \: sin50\degree \: sin70\degree = \dfrac{1}{8}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)