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Trigonometry> Prove that ~16cos24°.cos48°.cos96°.cos168...

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Prove that ~16cos24°.cos48°.cos96°.cos168° =1This question is from S.L.Loney Plane trigonometry part -I

Ayush Baghel , 6 Years ago

Grade 11

3 Answers

Deepak Kumar Shringi

Last Activity: 6 Years ago

Siddharth Singh

Last Activity: 6 Years ago

the dipole moment p induced in an atom of neon gas whe an electric field E is applied. , isP = \alpha\varepsilono E P = np = \varepsilono(K –1)E p = \varepsilono (K –1)E / N the volume of 1 m atoms of neon at N.T.P. is 22.4 litres. N = Na / V Hence on calculation p = 2.65 * 10 ^ –36 Cm

Soumendu Majumdar

Last Activity: 6 Years ago

Dear Student,

L.H.S = $4(2cos24^\circ cos96^\circ)(2cos48^{\circ}cos168^{\circ})$

= $4(cos120^{\circ}+cos72^{\circ})(cos216^{\circ}+cos120^{\circ})$

= $4(-sin30^{\circ}+sin18^{\circ})(-cos36^{\circ}-sin30^{\circ})$

Now,

$sin 18^{\circ}= (\sqrt5 -1)/4$

$cos 36^{\circ}= (\sqrt5 +1)/4$

so L.H.S =

$4(\sqrt5-3)(-3-\sqrt5)/16$

= $(9-5)/4$

= $1 =R.H.S$ [Proved]

Hope it helps!

with regards,

Soumendu

P.S-If you have any confusion in understanding the solution, feel free to ask, I’ll be glad to help you!

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