Evaluate

Geometry Level 2

Evaluate (in degrees)

sin ⁡ − 1 ( sin ⁡ 155 0 ∘ ) . \sin ^{-1} (\sin 1550^\circ ) .


The answer is 70.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Chew-Seong Cheong
Dec 10, 2015

The range of sin ⁡ − 1 x \sin^{-1} x is [ − 9 0 ∘ , 9 0 ∘ ] [-90^\circ, 90^\circ] , therefore,

sin ⁡ − 1 ( sin ⁡ 155 0 ∘ ) = sin ⁡ − 1 ( sin ⁡ ( 1550 mod 360 ) ∘ ) = sin ⁡ − 1 ( sin ⁡ 11 0 ∘ ) = sin ⁡ − 1 ( sin ⁡ ( 180 − 110 ) ∘ ) = sin ⁡ − 1 ( sin ⁡ 7 0 ∘ ) = 70 ∘ \begin{aligned} \sin^{-1}(\sin 1550^\circ) & = \sin^{-1}(\sin (1550 \text{ mod } 360)^\circ) \\ & = \sin^{-1}(\sin 110^\circ) \\ & = \sin^{-1}(\sin (180-110)^\circ) \\ & = \sin^{-1}(\sin 70^\circ) \\ & = \boxed{70}^\circ \end{aligned}

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Exactly. !!!!

A Former Brilliant Member - 5 years, 6 months ago

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