Bài 2 trang 154 SGK Đại số 10

Đề bài

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 \(\sin b =  \frac{2}{3},\) \(90^0< b < 180^0.\) 

Hướng dẫn giải

+) Với \(0 < \alpha  < \frac{\pi }{2}\) ta có: \(\sin \alpha >0, \, \, \cos \alpha >0.\)

+) Với \( \frac{\pi }{2} < \alpha  < \pi \) ta có: \(\sin \alpha >0, \, \, \cos \alpha < 0.\)

+) \(\sin^2 \alpha +\cos^2 \alpha =1. \)

+) \({\tan ^2}x + 1 = \frac{1}{{{{\cos }^2}x}}.\)

+)  \({\cot ^2}x + 1 = \frac{1}{{{{\sin }^2}x}}.\)

+) \(\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\)

+) \( \cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .\)

+) \( \cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta .\)

+) \( tan\left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.\)

+) \( tan\left( {\alpha - \beta } \right) = \frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}.\)

Lời giải chi tiết

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 \(\sinα > 0, \cosα < 0, \tanα < 0, \)\(\cotα < 0\)

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

\(tan(α -  \frac{\pi}{4}) =  \frac{\tan\alpha -\tan\frac{\pi}{4}}{1+\tan\alpha tan\frac{\pi}{4}}=\frac{-1-2\sqrt{2}}{1-2\sqrt{2}}\)

\(=\frac{2\sqrt{2}+1}{2\sqrt{2}-1}= \frac{{9 + 4\sqrt 2 }}{7}.\)

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

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

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

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

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

               \( =\frac{3}{5}\left ( -\frac{\sqrt{5}}{3} \right )-\frac{4}{5}.\frac{2}{3}=-\frac{3\sqrt{5}+8}{15}\) 

\(\eqalign{
& \sin(a - b) = \sin a\cos b - \cos a\sin b \cr
& = {4 \over 5}.\left( { - {{\sqrt 5 } \over 3}} \right) - {3 \over 5}.{2 \over 3} \cr&= - {{4\sqrt 5 + 6} \over {15}} \cr} \)