Question

cos cube 10+ cos cube 110+ cos cube130

Answer 1

Hi Dear !


cos³10 + cos³110 + cos³130 

we know , 

Cos³ x + Cos ³ ( 120-x )+Cos³ ( 120+ x ) = (3/4)Cos 3x

here x = 10 

now , 


= Cos10+ Cos³ (120-10)+ Cos³ (120+10)

= ( 3/4) Cos (3×10)

= (3/4)Cos30°

= (3/4)×(√3/2)

= (3√3)/8

Answer 2

So,The answer is $\frac{3\sqrt{3} }{8}$

Step-by-step explanation:

Given;

$cos^{3} 10+cos^{3}110+cos^{3}130$

We know,

$cos^{3}x+cos^{3}(120-x)+cos^{3}(120+x)=\frac{3}{4} cos3x$

∴$cos^{3} 10+cos^{3}110+cos^{3}130$

$=cos^{3} 10+cos^{3}(120-10)+cos^{3}(120+10)$

$=\frac{3}{4}cos(3\times10)$

$=\frac{3}{4}cos(30)$

$=\frac{3}{4}\times \frac{\sqrt{3} }{2}$

$=\frac{3\sqrt{3} }{8}$

∴ The answer is $\frac{3\sqrt{3} }{8}$