Tính cos 225 độ, sin 240 độ, cot (-15 độ), tan 75 độ

Giải Toán 10 Bài 3: Công thức lượng giác

Video Giải Bài 1 trang 153 Toán lớp 10 Đại số

Bài 1 trang 153 Toán lớp 10 Đại số:

Lời giải:

a)

+)

$\cos 225°=\cos \left(180°+45°\right)=-\cos 45°=-\frac{\sqrt{2}}{2}$

+) $\sin 240°=\sin \left(180°+60°\right)$

$=-\sin 60°=-\frac{\sqrt{3}}{2}$

+)

$\cot \left(-15°\right)=-\cot 15°=-\cot \left(90°-75°\right)$

$=-\tan 75°=-\tan \left(30°+45°\right)$

$=\frac{-\tan 30°-\tan 45°}{1-\tan 30°\tan 45°}$

$=\frac{-\frac{1}{\sqrt{3}}-1}{1-\frac{1}{\sqrt{3}}}=-\frac{\sqrt{3}+1}{\sqrt{3}-1}=-\frac{{(\sqrt{3}+1)}^2}{2}$

$=-2-\sqrt{3}$

+) $\tan 75°=\tan \left(90°-15°\right)$

$=\cot 15°$$=-\cot \left(-15°\right)$

$=-(-2-\sqrt{3})=2+\sqrt{3}$

b)

+) $\sin \frac{7\pi }{12}=\sin \left(\frac{\pi }{3}+\frac{\pi }{4}\right)$

$=\sin \frac{\pi }{3}\cos \frac{\pi }{4}+\cos \frac{\pi }{3}\sin \frac{\pi }{4}$

$=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}+\frac{1}{2}\cdot \frac{\sqrt{2}}{2}$

$=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}$

+) $\cos \left(-\frac{\pi }{12}\right)=\cos \left(\frac{\pi }{4}-\frac{\pi }{3}\right)$

$=\cos \frac{\pi }{4}\cos \frac{\pi }{3}+\sin \frac{\pi }{3}\sin \frac{\pi }{4}$

$=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}$

$=\frac{\sqrt{2}+\sqrt{6}}{4}$

+) $\tan \left(\frac{13\pi }{12}\right)=\tan \left(\pi +\frac{\pi }{12}\right)$

$=\tan \frac{\pi }{12}=\tan \left(\frac{\pi }{3}-\frac{\pi }{4}\right)$

$=\frac{\tan \frac{\pi }{3}-\tan \frac{\pi }{4}}{1+\tan \frac{\pi }{3}\tan \frac{\pi }{4}}$

$=\frac{\sqrt{3}-1}{1+\sqrt{3}.1}$

$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$$=\frac{{(\sqrt{3}-1)}^2}{3-1}$

$=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$