Math  Calculus
posted by Anonymous on .
The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below.
 a × b ² + (a • b)² = a²b²
My work, LSH:
=  a × b ² + (a • b)²
= (absinθ)(absinθ) + (abcosθ)(abcosθ)
= (a²)(ab)(asinθ)(ab)(b²)(bsinθ)(a sinθ)(bsinθ)(sin²θ) + (a²)(ab)(acosθ)(ab)(b²)(bcosθ)(a cosθ)(bcosθ)(cos²θ)
= (a²)(ab)²(asinθ)²(b²)(bsinθ)²(sin²θ) + (a²)(ab)²(acosθ)²(b²)(bcosθ)²(cos²θ)
= (a²b²(ab)²) [(asinθ)²(bsinθ)²(sin²θ) + (acosθ)²(b²)(bcosθ)²(cos²θ)]
= (a²b²(ab)²) [(a²)(sin²θ)(b²)(sin²θ)(sin²θ) + (a²)(cos²θ)(b²)(b²)(cos²θ)(cos²θ)]
= (a²b²(ab)²) [(sin²θ)(sin²θ)(sin²θ) + (cos²θ)(cos²θ)(cos²θ)]
And now I don't know what else to do! Please help. Did I mess up somewhere in my steps? Or is it possible to common factor still?

My work, LSH:
=  a × b ² + (a • b)²
= (absinθ)(absinθ) + (abcosθ)(abcosθ)
You can write this as:
a^2b^2[sin^2(theta) + cos^2(theta)]
The result follows from the fact that
sin^2(theta) + cos^2(theta) = 1