Can you expandHi there,
I need some help with this answer ASAP, please and thank you!
Simplify the expression by using a double-angle formula:
View attachment 13240
(I am attaching a screen shot of the formula because I can't find how to type the symbols).
Thanks!
Katie
It is a difference in rounding, that is all.Thank you for that explanation!
I am stumped as to why the 8th decimal is different. What am I missing?
Thank you!It is a difference in rounding, that is all.
Got it, I was overthinking it, thank you! ??Consider a different equation altogether that's much simpler to solve: \(\displaystyle \frac{1}{3} \cdot \frac{1}{6}\). What would happen if we used approximations instead of the real value?
\(\displaystyle \frac{1}{3} \approx 0.3333\) and \(\displaystyle \frac{1}{6} \approx 0.1666\)
\(\displaystyle 0.3333 \cdot 0.1666 = 0.0555{\color{red}2778}\)
\(\displaystyle \frac{1}{18} = 0.0555{\color{red}5555}\)
Could we make the approximation tighter? What happens then?
\(\displaystyle \frac{1}{3} \approx 0.33333333\) and \(\displaystyle \frac{1}{6} \approx 0.16666666\)
\(\displaystyle 0.33333333 \cdot 0.16666666 = 0.05555555{\color{red}27777778}\)
\(\displaystyle \frac{1}{18} = 0.05555555{\color{red}55555555}\)
? What happens if we use the real, exact value?
\(\displaystyle \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18}\)