Question

prove that cos 10° cos 30° cos 50° cos 70° = 3/16

Answer 1

cos30.(cos50cos10).cos70

= root3/2 .1/2(2cos10cos50)cos70

= root3/2 .1/2 (cos60 + cos40 ).cos70.............[ bcoz 2cosxcosy = cos(x+y) + cos(x-y) ]

= root3/4 (cos60.cos70 + cos40cos70)

= root3/8 [ cos70 + 2cos40cos70 ]

= root3/8 [ cos70 + cos110 + cos30 ]

= root3/8 [ cos70 - cos70 + root3/2]

= root3/8 . root3/2

= 3/16

Answer 2

Answer and explanation:

To prove : $\cos10^\circ\cos30^\circ\cos 50^\circ\cos 70^\circ=\frac{3}{16}$

Proof :

Taking LHS,

$\cos10^\circ\cos30^\circ\cos 50^\circ\cos 70^\circ$

$=\cos30^\circ(\cos50^\circ\cos 10^\circ)\cos 70^\circ$

$=\frac{\sqrt3}{2}\times \frac{1}{2}(2\cos50^\circ\cos 10^\circ)\cos 70^\circ$

Applying identity, $2\cos A\cos B=\cos(A+B)+\cos(A-B)$

$=\frac{\sqrt3}{2}\times \frac{1}{2}(\cos(50+10)+\cos (50-10))\cos 70^\circ$

$=\frac{\sqrt3}{2}\times \frac{1}{2}(\cos(60)+\cos (40))\cos 70^\circ$

$=\frac{\sqrt3}{4}(\frac{1}{2}+\cos (40))\cos 70^\circ$

$=\frac{\sqrt3}{8}(\cos 70+2\cos (40)\cos 70^\circ)$

$=\frac{\sqrt3}{8}(\cos 70+\cos 110+\cos 30)$

$=\frac{\sqrt3}{8}(\cos 70-\cos 70+\frac{\sqrt3}{2})$

$=\frac{\sqrt3}{8}(\frac{\sqrt3}{2})$

$=\frac{\sqrt3}{16}$

= RHS

Hence proved.