Hey,
Anyone know if this can be simplified any further?
$\displaystyle [\sqrt{4(cos^2t) + 9(sin^2t)} ]^3$
Cheers
Wooo, according to you, it is correct to write something like $\displaystyle 5=\sqrt{25}=\sqrt{9+16}= \sqrt{3^2+4^2}=3+4=7$ ?!
The problem is that, in general, $\displaystyle \sqrt{a^2+b^2}$ does not equal $\displaystyle a+b$ : you can't say that $\displaystyle \sqrt{4 \cos^2t + 9\sin^2t}=2\cos t+3\sin t$ and you can't say that $\displaystyle \sqrt{3^2+4^2}=3+4$.
However, it is correct to write $\displaystyle \left[\sqrt{x}\right]^2=x$ or $\displaystyle \sqrt{x^2}=|x|$... but these two identities can't be used here.
$\displaystyle [2cost + 3sint]^3$
$\displaystyle 8cos^3t + 27sin^3t$
Again, in general, $\displaystyle (a+b)^3$ does not equal $\displaystyle a^3+b^3$. Remember that $\displaystyle (a+b)^3=(a+b)(a+b)(a+b)$. If you expand this, you'll see that $\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3$...