Giải câu 2 bài 3: Công thức lượng giác sgk Đại số 10 trang 154.

a) Vì \(0 < α <  \frac{\pi}{2}\) nên \(\sinα > 0, \cosα > 0\)

\(\cosα  =  \sqrt{1-\sin^{2}\alpha }=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\)

\(cos(α + \frac{\pi}{3}) \)

\(= \cosα\cos \frac{\pi }{3} - \sinα\sin \frac{\pi}{3}\)

\(=\frac{\sqrt{6}}{3}.\frac{1}{2}-\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{2}\)

\(=\frac{\sqrt{6}-3}{6}\)

b) Vì \( \frac{\pi}{2}< α < π\) nên \(\sinα > 0, \cosα < 0, \tanα < 0, \cotα < 0\)

\(\tanα = -\sqrt{\frac{1}{cos^{2}\alpha }-1}=-\sqrt{3^{3}-1} = -2\sqrt2\)

\(tan(α -  \frac{\pi}{4}) \)

\(=  \frac{\tan\alpha -\tan\frac{\pi}{4}}{1+\tan\alpha tan\frac{\pi}{4}}\)

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

c) Vì \(0^0< a < 90^0\Rightarrow  \sin a > 0, \cos a > 0\)

        \(90^0< b < 180^0\Rightarrow \sin b > 0, \cos b < 0\)

\(\cos a =  \sqrt{1-sin^{2}a}=\sqrt{1-\left ( \frac{4}{5} \right )^{2}}=\frac{3}{5}\)

\(\cos b =  -\sqrt{1-sin^{2}a}=-\sqrt{1-\left ( \frac{2}{3} \right )^{2}}=-\frac{\sqrt{5}}{3}\)

Ta lại có:

  • \(\cos(a + b) = \cos a\cos b - \sin a\sin b\)

\(=\frac{3}{5}\left ( -\frac{\sqrt{5}}{3} \right )-\frac{4}{5}.\frac{2}{3}\)

\(=-\frac{3\sqrt{5}+8}{15}\) 

  • \(\sin(a - b) = \sin a\cos b - \cos a\sin b \)

\(= {4 \over 5}.\left( { - {{\sqrt 5 } \over 3}} \right) - {3 \over 5}.{2 \over 3} \)

\(= - {{4\sqrt 5 + 6} \over {15}} \)