Question

[Sách Giáo Khoa]

Tính :

a) \(\cos225^0;\sin240^0;\cot\left(-15^0\right);\tan75^0\)

b) \(\sin\dfrac{7\pi}{15};\cos\left(-\dfrac{\pi}{12}\right);\tan\dfrac{13\pi}{12}\)

Answer 1

[Hoang Hung Quan]

a)

\(\cos225^0=\cos\left(180^0+45^0\right)=-\cos45^0=-\dfrac{\sqrt{2}}{2}\)

\(\sin240^0=\sin\left(180^0+60^0\right)=-\sin60^0=-\dfrac{\sqrt{3}}{2}\)

\(\cos\left(-15^0\right)=-\cot15^0=-\tan75^0=-\tan\left(30^0+45^0\right)\)

\(=\dfrac{-\tan30^0-\tan45^0}{1-\tan30^0\tan45^0}=\dfrac{-\dfrac{1}{\sqrt{3}}-1}{1-\dfrac{1}{\sqrt{3}}}=-\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)

\(=-\dfrac{\left(\sqrt{3}+1\right)^2}{2}=-2-\sqrt{3}\)

\(\tan75^0=\cot15^0=2+\sqrt{3}\)

b)

\(\sin\dfrac{7\pi}{12}=\sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}+\cos\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)

\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

\(\cos\left(-\dfrac{\pi}{12}\right)=\cos\left(\dfrac{\pi}{4}-\dfrac{\pi}{3}\right)=\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{3}+\sin\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)

\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\)

\(\tan\dfrac{13\pi}{12}=\tan\left(\pi+\dfrac{\pi}{12}\right)=\tan\dfrac{\pi}{12}=\tan\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)\)

\(=\dfrac{\tan\dfrac{\pi}{3}-\tan\dfrac{\pi}{4}}{1+\tan\dfrac{\pi}{3}\tan\dfrac{\pi}{4}}=\dfrac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3}\)