Question

coscube10 +coscube110+coscube130=√3x, then 8x=?​

Answer 1

Answer:

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

4.3

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Comments Report

guptasingh4564 Ambitious

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}