HSF.TF.B.6 math practice. Learn by doing.

HSF.TF.B.6 practice covers understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed. Work through 8 free questions with answers and explanations, then continue in the related math games.

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What is arccos⁡(12)\arccos\left(\dfrac{1}{2}\right), in degrees?

cos⁡60∘=12\cos 60^{\circ} = \dfrac{1}{2} and 60∘60^{\circ} lies in [0∘,180∘][0^{\circ}, 180^{\circ}], so arccos⁡(12)=60∘\arccos\left(\dfrac{1}{2}\right) = 60^{\circ}.

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What is arcsin⁡(12)\arcsin\left(\dfrac{1}{2}\right), in degrees?

sin⁡30∘=12\sin 30^{\circ} = \dfrac{1}{2} and 30∘30^{\circ} lies in [−90∘,90∘][-90^{\circ}, 90^{\circ}], so arcsin⁡(12)=30∘\arcsin\left(\dfrac{1}{2}\right) = 30^{\circ}.

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What is arctan⁡(1)\arctan(1), in degrees?

tan⁡45∘=1\tan 45^{\circ} = 1 and 45∘45^{\circ} lies in arctangent's principal range, so arctan⁡(1)=45∘\arctan(1) = 45^{\circ}.

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What is the domain of arcsin⁡(x)\arcsin(x)?

The domain of arcsin⁡(x)\arcsin(x) is [−1, 1][-1,\, 1].

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What is arctan⁡(1)\arctan(1)?

tan⁡ ⁣(π4)=1\tan\!\left(\dfrac{\pi}{4}\right) = 1, so arctan⁡(1)=π4\arctan(1) = \dfrac{\pi}{4}.

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What is arccos⁡(0)\arccos(0), in degrees?

cos⁡90∘=0\cos 90^{\circ} = 0 and 90∘90^{\circ} lies in [0∘,180∘][0^{\circ}, 180^{\circ}], so arccos⁡(0)=90∘\arccos(0) = 90^{\circ}.

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What is arcsin⁡(1)\arcsin(1), in degrees?

sin⁡90∘=1\sin 90^{\circ} = 1, the top of arcsine's range [−90∘,90∘][-90^{\circ}, 90^{\circ}], so arcsin⁡(1)=90∘\arcsin(1) = 90^{\circ}.

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What is arctan⁡(0)\arctan(0), in degrees?

tan⁡0∘=0\tan 0^{\circ} = 0, so arctan⁡(0)=0∘\arctan(0) = 0^{\circ}.

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