# Thread: Rearranging equation

1. ## Rearranging equation

Can you rearrange an equation like what i have done?

$\displaystyle a^2cos^2\theta a^2sin^2\theta d\theta = a^4cos^2\theta sin^2\theta d\theta$

2. Originally Posted by Karl Harder
Can you rearrange an equation like what i have done?

$\displaystyle a^2cos^2\theta a^2sin^2\theta d\theta = a^4cos^2\theta sin^2\theta d\theta$
as long as each $\displaystyle a$ is a distinct constant factor, you sure can.

you can also do this ...

$\displaystyle a^4\cos^2{\theta} \sin^2{\theta} \, d\theta =$

$\displaystyle \frac{a^4}{4} \sin^2(2\theta) \, d\theta =$

$\displaystyle \frac{a^4}{8}[1 - \cos(4\theta)] \, d\theta$

3. Originally Posted by skeeter
as long as each $\displaystyle a$ is a distinct constant factor, you sure can.

you can also do this ...

$\displaystyle a^4\cos^2{\theta} \sin^2{\theta} \, d\theta =$

$\displaystyle \frac{a^4}{4} \sin^2(2\theta) \, d\theta =$

$\displaystyle \frac{a^4}{8}[1 - \cos(4\theta)] \, d\theta$
What maths principles did you use to get these answers. I have been trying to use the half angle formulas is this correct?

4. Originally Posted by Karl Harder
What maths principles did you use to get these answers. I have been trying to use the half angle formulas is this correct?
I used a double angle formula and a power reduction identity (derived from a double angle formula) ...

$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$

$\displaystyle \sin^2{u} = \frac{1 - \cos(2u)}{2}$

5. I'm still stuck on how you get this using the double angle formula?

$\displaystyle a^4\cos^2{\theta} \sin^2{\theta} \, d\theta = \frac{a^4}{4} \sin^2(2\theta) \, d\theta$

6. Originally Posted by Karl Harder
I'm still stuck on how you get this using the double angle formula?

$\displaystyle a^4\cos^2{\theta} \sin^2{\theta} \, d\theta = \frac{a^4}{4} \sin^2(2\theta) \, d\theta$
never mind i have found out that i can get the answer by using the proudct formula

7. Originally Posted by Karl Harder
I'm still stuck on how you get this using the double angle formula?

$\displaystyle a^4\cos^2{\theta} \sin^2{\theta} \, d\theta = \frac{a^4}{4} \sin^2(2\theta) \, d\theta$

... never mind i have found out that i can get the answer by using the proudct formula
product formula? ...

$\displaystyle a^4 \cos^2{\theta} \sin^2{\theta}$

$\displaystyle \frac{a^4}{4} \cdot 4\cos^2{\theta} \sin^2{\theta}$

$\displaystyle \frac{a^4}{4} (2\cos{\theta} \sin{\theta})^2$

$\displaystyle \frac{a^4}{4} \sin^2(2\theta)$