Oh my!

Geometry Level 3

$\displaystyle \sum_{x = 0} ^{100} \lfloor \sin (x ^{\circ}) \rfloor = \, ?$

Give your answer to 2 decimal places.

The answer is 1.00.

3 solutions

Kay Xspre
Feb 1, 2016

For $0\leq x < \frac{\pi}{2}$ , we will get $0\leq sin(x)<1$ , and $\lfloor sin(x)\rfloor = 0$
Since $sin(\frac{\pi}{2}+x)=cos(x)$ , for $0 < x \leq \frac{\pi}{18}$ , $0 < sin(\frac{\pi}{2}+x) < 1$ , and $\lfloor sin(\frac{\pi}{2}+x) \rfloor = 0$

From the properties above we can conclude that all except $sin(\frac{\pi}{2})$ will be zero. As $sin(\frac{\pi}{2}) = 1$ , then the sum of the above series is 1.

Colin Carmody
Feb 1, 2016

Sin (90) is the only value of sin (x) that is 1 or greater. So all other values will round down to zero. So you have 0+0+0...+1+0+0...=1.00

Kushagra Sahni
Feb 2, 2016

If there is involvement of the floor function why does it ask us to give the answer till two decimal places?

Just to confuse us and think us again.................

Abhisek Mohanty - 5 years, 4 months ago

No one will be confused I am sure, provided that the person knows about the floor function.

Kushagra Sahni - 5 years, 4 months ago

Asking for $\displaystyle\sum_{k=0}^{100} \lfloor \cos(k^{\circ}) \rfloor$ might be a better test of people's understanding of the floor function, (the answer of course being $-9$ ).

Brian Charlesworth - 5 years, 4 months ago

Oh my!

The answer is 1.00.

3 solutions

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