Trig Distributive Problem - # 3

Jason76

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Oct 19, 2012
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(π4)[[4sinπ4]+[4sinπ2]+[4sin3π4]+[4sinπ]+[4sin5π4]+[4sin3π2]]\displaystyle (\dfrac{\pi }{4})[[4 \sin \dfrac{\pi}{4}] + [4 \sin \dfrac{\pi }{2}] + [4 \sin \dfrac{3\pi }{4}] + [4 \sin \pi] + [4 \sin \dfrac{5\pi }{4}] + [4 \sin \dfrac{3\pi }{2}]] :confused: What does this come out to?

My attempt:

(π4)[[4(.707106781)]+[4(1)]+[4(.70710678118)]+[4(0)]+[4(.7071067811)]+[4(1)]]\displaystyle (\dfrac{\pi }{4})[[4 (.707106781)] + [4 (1)] + [4(.70710678118)] + [4 (0)] + [4 (-.7071067811)] + [4(-1)]]
 
Last edited:
(π4)[[4sinπ4]+[4sinπ2]+[4sin3π4]+[4sinπ]+[4sin5π4]+[4sin3π2]]\displaystyle (\dfrac{\pi }{4})[[4 \sin \dfrac{\pi}{4}] + [4 \sin \dfrac{\pi }{2}] + [4 \sin \dfrac{3\pi }{4}] + [4 \sin \pi] + [4 \sin \dfrac{5\pi }{4}] + [4 \sin \dfrac{3\pi }{2}]] :confused: What does this come out to?

My attempt:

(π4)[[4(.707106781)]+[4(1)]+[4(.70710678118)]+[4(0)]+[4(.7071067811)]+[4(1)]]\displaystyle (\dfrac{\pi }{4})[[4 (.707106781)] + [4 (1)] + [4(.70710678118)] + [4 (0)] + [4 (-.7071067811)] + [4(-1)]]

Now finish it.....
 
(π4)[[4sinπ4]+[4sinπ2]+[4sin3π4]+[4sinπ]+[4sin5π4]+[4sin3π2]]\displaystyle (\dfrac{\pi }{4})[[4 \sin \dfrac{\pi}{4}] + [4 \sin \dfrac{\pi }{2}] + [4 \sin \dfrac{3\pi }{4}] + [4 \sin \pi] + [4 \sin \dfrac{5\pi }{4}] + [4 \sin \dfrac{3\pi }{2}]] :confused: What does this come out to?

My attempt:

(π4)[[4(.707106781)]+[4(1)]+[4(.70710678118)]+[4(0)]+[4(.7071067811)]+[4(1)]]\displaystyle (\dfrac{\pi }{4})[[4 (.707106781)] + [4 (1)] + [4(.70710678118)] + [4 (0)] + [4 (-.7071067811)] + [4(-1)]]
It might be better to write it as
π[22+1+22+0221]\displaystyle \pi\left[\dfrac{\sqrt{2}}{2}+ 1+ \dfrac{\sqrt{2}}{2}+ 0- \dfrac{\sqrt{2}}{2}- 1\right]
where I have factored the "4" out of the sum, canceled it with the 4 in the denominator, and written the exact values for the sines.
 
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