Hm...Isaac's Trigonometry.

Geometry Level 4

Try not to use a calculator for this one:

Find the value of sin ⁡ ( 2 ° × 10 − 11 ) − sin ⁡ ( 2 ° × 10 − 12 ) \sin { (2°\times{ 10 }^{ -11 } } )-\sin { (2°\times{ 10 }^{ -12 } } )

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None of the Above π × 10 − 13 \pi\times{10}^{-13} π × 10 − 12 \pi\times{10}^{-12} 0

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Julian Poon by Julian Poon , 20, Singapore

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2 solutions

Julian Poon
Nov 5, 2014

sin ⁡ x ° \sin{x°} When x ° x° is very small, sin ⁡ x ° \sin{x°} can be closely approximated to be equal to x ° x° in radians. So, sin ⁡ ( 2 ° × 10 − 11 ) − sin ⁡ ( 2 ° × 10 − 12 ) ≈ 2 × 10 − 11 × π 180 − 2 × 10 − 12 × π 180 \sin{(2°\times{10}^{-11})}-\sin{(2°\times{10}^{-12})}\approx \frac{2\times{10}^{-11}\times\pi}{180}-\frac{2\times{10}^{-12}\times\pi}{180} = π × 10 − 13 =\pi\times{10}^{-13} So, even though it can be very closely approximated to be π × 10 − 13 \pi\times{10}^{-13} , it isn't the answer. Therefore, the answer is None of the Above \boxed{\textbf{None of the Above}}

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Luke Zhang
Feb 1, 2015

Very good. Good questions Julian

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