Question

(iv) cos 40°cos 100°cos 160º = 1/8
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Answer 1

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

Consider,

$\rm :\longmapsto\:cos40\degree cos100\degree cos160\degree$

$\rm \: = \:cos40\degree cos(180\degree - 80\degree )cos(180\degree - 20\degree )$

We know,

$\boxed{ \tt{ \: cos(180\degree - x) = - \: cosx \: }}$

$\rm \: = \:cos40\degree [ - cos80\degree ] \: [ - cos20\degree ]$

$\rm \: = \:cos20\degree cos40\degree cos80\degree$

$\rm \: = \:cos20\degree cos(60\degree - 20\degree) cos(60\degree + 20\degree )$

We know,

$\boxed{ \tt{ \: cosxcos(60\degree - x)cos(60\degree + x) = \frac{1}{4}cos3x \: }}$

So, using this identity, we get

$\rm \: = \:\dfrac{1}{4} \: cos[3 \times 20\degree ]$

$\rm \: = \:\dfrac{1}{4} \: cos60\degree$

$\rm \: = \:\dfrac{1}{4} \times \dfrac{1}{2}$

$\rm \: = \:\dfrac{1}{8}$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: cos40\degree cos100\degree cos160\degree = \frac{1}{8} \: }}$

More Identities to know :-

$\boxed{ \tt{ \: sinxsin(60\degree - x)sin(60\degree + x) = \frac{1}{4}sin3x \: }}$

$\boxed{ \tt{ \: tanxtan(60\degree - x)tan(60\degree + x) = tan3x \: }}$

$\boxed{ \tt{ \: cosx \: cos2x \: cos {2}^{2}x - - cos {2}^{n}x = \frac{sin {2}^{n + 1} x}{ {2}^{n + 1}sinx}}}$