Question

Prove that Sin^6 theta+ Cos^6 theta + 3Cos² theta Sin²theta= 1

Answer 1

Hiii friend,


Sin^6 theta + Cos^6 theta + 3 × Cos² theta × Sin² theta = 1

We have,

LHS = Sin^6 theta + Cos^6 theta + 3 × Cos²theta × Sin²theta

We know that,

Sin² theta +Cos² theta = 1


Therefore,


(Sin²theta + Cos²theta)³ = (1)³ = 1


=> (Sin²theta)³ + (Cos²thet)³ + 3 × Sin²theta × Cos²theta (Sin²theta+Cos²theta) = 1



=> Sin^6 theta + Cos^6 theta + 3 × Sin²theta × Cos²theta × 1 = 1


=> Sin^6 theta + Cos^6 theta + 3 × Sin²theta×Cos²theta = 1


Hence,


LHS = RHS...... PROVED...


HOPE IT WILL HELP YOU...... :-)

Answer 2

Hi ,

Here I am using A instead of theta.


LHS = sin^6 A + cos^6A+3cos²Asin²A

= ( sin² A)³ + ( cos² A )³+3cos²Asin²A

= ( sin² A+cos² A )³ -

3sin²Acos²A(sin² A+cos²A)+3sin²Acos²A

*********************************
We know the algebraic identity ,

1 ) a³ + b³ = ( a + b )³ - 3ab( a + b )

Here ,

a = sin² A ,

b = cos² A

2 ) sin² A + cos² A = 1

*********************************

= 1 - 3cos²Asin²A + 3cos²Asin²A

= 1

= RHS

Hence proved.

I hope this helps you.

: )