Question

Đã trả lời bởi chuyên gia

Answer 1

b) Ta có:

– 555° = $\frac{π . (- 555 °)}{180 °} = - \frac{37 π}{12} = - (3 π + \frac{π}{12})$ rad.

Khi đó:

+) $cos (- 555 °) = cos (3 π + \frac{π}{12}) = - cos (\frac{π}{12}) = - cos (\frac{π}{3} - \frac{π}{4})$

$= - (cos \frac{π}{3} cos \frac{π}{4} + \sin \frac{π}{3} \sin \frac{π}{4}) = - (\frac{1}{2} . \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} . \frac{\sqrt{2}}{2}) = - \frac{\sqrt{2} + \sqrt{6}}{4}$ .

+) $\sin (- 555 °) = \sin (3 π + \frac{π}{12}) = \sin (\frac{π}{12}) = \sin (\frac{π}{3} - \frac{π}{4})$

$= \sin \frac{π}{3} cos \frac{π}{4} - cos \frac{π}{3} \sin \frac{π}{4} = \frac{\sqrt{3}}{2} . \frac{\sqrt{2}}{2} - \frac{1}{2} . \frac{\sqrt{2}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$ .

+) $\tan (- 555 °) = \frac{\sin (- 555 °)}{cos (- 555 °)} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{- \frac{\sqrt{6} + \sqrt{2}}{4}} = - 2 + \sqrt{3}$ .

+) $\cot (- 555 °) = \frac{1}{- 2 + \sqrt{3}} = - 2 - \sqrt{3}$ .

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Answer 2