Giải Bài 1 trang 23 SGK Toán 11 Chân trời sáng tạo tập 1 - CTST

Phương pháp giải

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Lời giải chi tiết

\(sin(\frac{5\pi }{12}) = sin(\frac{\pi }{6}+\frac{\pi }{4}) \)\(= sin(\frac{\pi }{6}).cos(\frac{\pi }{4})+cos(\frac{\pi }{6}).sin(\frac{\pi }{4}) \)\(= \frac{1}{2}.\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2} \)\(= \frac{\sqrt{2}+\sqrt{6}}{4}\)

\(cos(\frac{5\pi }{12}) = cos(\frac{\pi }{6}+\frac{\pi }{4}) \)\(= cos(\frac{\pi }{6}).cos(\frac{\pi }{4})-sin(\frac{\pi }{6}).sin(\frac{\pi }{4}) \)\(= \frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2} \)\(= \frac{\sqrt{6}-\sqrt{2}}{4}\)

\(tan(\frac{5\pi }{12}) = \frac{sin(\frac{5\pi }{12})}{cos(\frac{5\pi }{12})} \)\(= \frac{\frac{\sqrt{2}+\sqrt{6}}{4}}{\frac{\sqrt{6}-\sqrt{2}}{4}}\)\(=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\)

\(sin(-555^{o}) = sin(720^{o}-555^{o}) = sin165^{o} \)\(= sin(180^{o}-165^{o}) = sin15^{o} = sin(45^{o}-30^{o})= sin(45^{o}).cos(30^{o})-cos(45^{o}).sin(30^{o})\)\( = \frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}.\frac{1}{2} \)\(= \frac{\sqrt{6}-\sqrt{2}}{4}\)

\(cos(-555^{o}) = cos(720^{o}-555^{o}) = cos165^{o}\)\( = -cos(180^{o}-165^{o}) = -cos15^{o} \)\(= -cos(45^{o}-30^{o})\)\(= -cos(45^{o}).cos(30^{o})-sin(45^{o}).sin(30^{o})\)\( = -\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}.\frac{1}{2} \)\(= -\frac{\sqrt{6}+\sqrt{2}}{4}\)

\(tan(-555^{o}) = \frac{sin(-555^{o})}{cos(-555^{o})}\)\( = \frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{-\frac{\sqrt{6}+\sqrt{2}}{4}}\)\(=\frac{-\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}\)

-- Mod Toán 11 HỌC247