Question

Need help simplifying sin^3 (3x^2) cos^3 (3x^2)

I am looking on Paul's Math Notes and the two simplifications he has after that are (1/2 sin(2(3x^2))^3

and in the last form it has 1/8 sin^3 (6x^2) .

Answer 1

I know it is based on sin(2t) = 2sin(t)cos(t) but I cant figure out how those fractions get there...

Answer 2

\[\sin^3 (3x^2) \cos^3 (3x^2)\]\[=[\sin (3x^2) \cos (3x^2)]^3\]\[=[{1 \over 2} \times 2 \sin (3x^2) \cos (3x^2)]^3\]\[=[{1 \over 2} \times \sin 2(3x^2)]^3\]\[=[{1 \over 2} \times \sin (6x^2)]^3\]\[={1 \over 8} \times \sin^3 (6x^2)\]

Let me know if you have any questions!

Answer 3

Oh so, the 1/2 goes in there because the equation does not actually start out with 2sin(x)cos(x). So we just put that 2 in there and then cancel it out with the 1/2?

Answer 4

Yup - you're just going backwards. You're right about using the sin(2x)=2sinxcosx identity. The expression gives as the sinxcosx part already, and if we want to add in the 2 and use the identity, we also need to introduce ½ so that the expression stays the same!

Answer 5

Part of your response is coming up as wingding symbols, maybe something you copied and pasted. But I understand now. Thank you for your response!

Answer 6

very cool

Answer 7

Just refresh the page - that happens sometimes with funny characters!

And glad I could help!