I got tricked by this

Geometry Level 3

Compute

sin ( cos 1 ( sin ( sin 1 ( cos ( sin 1 ( 0.3 ) ) ) ) ) ) . \sin(\cos^{-1}(\sin(\sin^{-1}(\cos(\sin^{-1}(-0.3)))))) .

0.3 0.3 1 0. 3 2 \sqrt{1-0.3^2} 0.3 -0.3 1 0. 3 2 -\sqrt{1-0.3^2}

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2 solutions

Note first that cos ( sin 1 ( 0.3 ) ) = cos ( sin 1 ( 0.3 ) ) = cos ( sin 1 ( 0.3 ) ) . \cos(\sin^{-1}(-0.3)) = \cos(-\sin^{-1}(0.3)) = \cos(\sin^{-1}(0.3)).

Next, working from the middle of the expression outward, we then note that (i): sin ( sin 1 ( y ) ) = y \sin(\sin^{-1}(y)) = y for 1 y 1. -1 \le y \le 1. As y = cos ( sin 1 ( 0.3 ) ) y = \cos(\sin^{-1}(0.3)) does lie in this interval, the given expression simplifies to

sin ( cos 1 ( cos ( sin 1 ( 0.3 ) ) ) . \sin(\cos^{-1}(\cos(\sin^{-1}(0.3))).

Now cos 1 ( cos ( y ) ) = y \cos^{-1}(\cos(y)) = y for 0 y π . 0 \le y \le \pi. Since y = sin 1 ( 0.3 ) y = \sin^{-1}(0.3) does lie in this interval, the expression further simplifies to

sin ( sin 1 ( 0.3 ) ) = 0.3 \sin(\sin^{-1}(0.3)) = \boxed{0.3}

due to statement (i) outlined above.

@Trevor Arashiro Did you go with 0.3 -0.3 ? I always have to be very deliberate with expressions like these, hence my exposition being somewhat excessive. :)

Brian Charlesworth - 5 years, 7 months ago

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Hahah, yes, I did. I simply glanced at every thing and since this was an easy contest, time was the only factor, so I never took a close look.

Trevor Arashiro - 5 years, 7 months ago
Aravind M
Dec 11, 2015

OMG! I did not notice there was a minus b4 0.3.... ;) Well you'll get the same answer if you proceed with 0.3. :D

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