Nội dung bài giảng
Bài 2. Tính
a) \(\cos(α + \frac{\pi}{3}\)), biết \(\sinα = \frac{1}{\sqrt{3}}\) và \(0 < α < \frac{\pi }{2}\).
b) \(\tan(α - \frac{\pi }{4}\)), biết \(\cosα = -\frac{1}{3}\) và \( \frac{\pi }{2} < α < π\)
c) \(\cos(a + b), \sin(a - b)\) biết \(\sin a = \frac{4}{5}\), \(0^0< a < 90^0\) và \(\sin b = \frac{2}{3}\), \(90^0< b < 180^0\)
Giải
a) Do \(0 < α < \frac{\pi}{2}\) 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) Do \( \frac{\pi}{2}< α < π\) nên \(\sinα > 0, \cosα < 0, \tanα < 0, \cotα < 0\)
\(\tanα = -\sqrt{\frac{1}{cos^{2}\alpha }-1}=-\sqrt{3^{3}-1} = -2\sqrt2\)
\(tan(α - \frac{\pi}{4}) = \frac{\tan\alpha -\tan\frac{\pi}{4}}{1+\tan\alpha tan\frac{\pi}{4}}=\frac{-1-2\sqrt{2}}{1-2\sqrt{2}}=\frac{2\sqrt{2}+1}{2\sqrt4{2}-1}\)
c) \(0^0< a < 90^0\Rightarrow \sin a > 0, \cos a > 0\)
\(90^0< b < 180^0\Rightarrow \sin b > 0, \cos b < 0\)
\(\cos a = \sqrt{1-sin^{2}a}=\sqrt{1-\left ( \frac{4}{5} \right )^{2}}=\frac{3}{5}\)
\(\cos b = -\sqrt{1-sin^{2}a}=-\sqrt{1-\left ( \frac{2}{3} \right )^{2}}=-\frac{\sqrt{5}}{3}\)
\(\cos(a + b) = \cos a\cos b - \sin a\sin b\)
\( =\frac{3}{5}\left ( -\frac{\sqrt{5}}{3} \right )-\frac{4}{5}.\frac{2}{3}=-\frac{3\sqrt{5}+8}{15}\)
\(\eqalign{
& \sin(a - b) = \sin a\cos b - \cos a\sin b \cr
& = {4 \over 5}.\left( { - {{\sqrt 5 } \over 3}} \right) - {3 \over 5}.{2 \over 3} = - {{4\sqrt 5 + 6} \over {15}} \cr} \)