Trig Distributive Problem - # 4

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
\(\displaystyle (\dfrac{\pi }{4})[[\dfrac{\pi}{8}] + [\dfrac{3\pi }{8}] + [\dfrac{5\pi }{8}] + [\dfrac{7\pi }{8}] + [\dfrac{9\pi }{8}] + [\dfrac{11\pi }{8}]]\) :confused: What does this come out to?

My attempt:

\(\displaystyle [\dfrac{\pi}{32}] + [\dfrac{3\pi }{32}] + [\dfrac{5\pi }{32}] + [\dfrac{5\pi }{32}] + [\dfrac{7\pi }{32}] + [\dfrac{9\pi }{32}] + [\dfrac{11\pi }{32}] ]\)

\(\displaystyle = \dfrac{36\pi}{32} = \dfrac{9\pi}{8}\)
 
Last edited:
\(\displaystyle (\dfrac{\pi }{4})[[\dfrac{\pi}{8}] + [\dfrac{3\pi }{8}] + [\dfrac{5\pi }{8}] + [\dfrac{7\pi }{8}] + [\dfrac{9\pi }{8}] + [\dfrac{11\pi }{8}]]\) :confused: What does this come out to?

My attempt:

\(\displaystyle [\dfrac{\pi}{32}] + [\dfrac{3\pi }{32}] + [\dfrac{5\pi }{32}] + [\dfrac{5\pi }{32}] + [\dfrac{7\pi }{32}] + [\dfrac{9\pi }{32}] + [\dfrac{11\pi }{32}] ]\)

\(\displaystyle = \dfrac{36\pi}{32} = \dfrac{9\pi}{8}\)

\(\displaystyle (\dfrac{\pi }{4}) * [[\dfrac{\pi}{8}] + [\dfrac{3\pi }{8}] + [\dfrac{5\pi }{8}] + [\dfrac{7\pi }{8}] + [\dfrac{9\pi }{8}] + [\dfrac{11\pi }{8}]]\)

\(\displaystyle = (\dfrac{\pi^2 }{4}) * [[\dfrac{1}{8}] + [\dfrac{3}{8}] + [\dfrac{5}{8}] + [\dfrac{7}{8}] + [\dfrac{9}{8}] + [\dfrac{11}{8}]]\)

\(\displaystyle = (\dfrac{\pi^2 }{4}) * [\dfrac{1+3+5+7+9+11}{8}] \)

\(\displaystyle = \dfrac{9\pi^2 }{8}\)
 
Top