Bài 1 trang 153 SGK Đại số 10

LG a

$\cos {225^0},\, \sin {240^0}, \, \cot( - {15^0}), \, \tan{75^0}$;

Phương pháp giải:

Áp dụng các công thức:

$\begin{array}{l} + )\;\cos \left( {\alpha + {{180}^0}} \right) = - \cos \alpha .\\ + )\;\sin\left( {\alpha + {{180}^0}} \right) = - \sin \alpha .\\ + )\;\cot \left( { - \alpha } \right) = - \cot \alpha .\\ + )\;\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\\ + )\;\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .\\ + )\;\tan\left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}. \end{array}$

Lời giải chi tiết:

$\cos{225^0} = \cos({180^0} +{45^0})$ $= - \cos{45^{0}}$ $= -\dfrac{\sqrt{2}}{2}$

+) $\sin{240^0} = \sin({180^0} +{60^0})$

 $= - \sin{60^0}=  -\dfrac{\sqrt{3}}{2}$

+) $\cot( - {15^0})= - \cot{15^0}$ $=  - \cot \left( {{{90}^0} - {{75}^0}} \right)$

$=  - \tan{75^0} =- \tan({30^0} +{45^0})$

$=\dfrac{-\tan30^{0}-\tan45^{0}}{1-\tan30^{0}\tan45^{0}}$

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

$= -2 - \sqrt 3$

+) $\tan 75^0 = \tan \left( {{{90}^0} - {{15}^0}} \right)$ $= \cot 15^0=-\cot (-15^0)$ $=-(-2 - \sqrt 3)= 2 + \sqrt3$

LG b

$\sin \dfrac{7\pi}{12},$ $\cos \left ( -\dfrac{\pi}{12} \right ),$ $\tan\left ( \dfrac{13\pi}{12} \right )$

Lời giải chi tiết:

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

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

$= \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} + \dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2}$ $= \dfrac{{\sqrt 6 }}{4} + \dfrac{{\sqrt 2 }}{4} = \dfrac{{\sqrt 6  + \sqrt 2 }}{4}$

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

$= \cos \dfrac{\pi }{4}\cos\dfrac{\pi }{3} + \sin \dfrac{\pi }{3}\sin \dfrac{\pi }{4}$ $=\dfrac{{\sqrt 2 }}{2} . \dfrac{1}{2}+ \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2}$ $= \dfrac{{\sqrt 2 }}{4} + \dfrac{{\sqrt 6 }}{4} = \dfrac{{\sqrt 2  + \sqrt 6 }}{4}$

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

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

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

$= \dfrac{{\sqrt 3  - 1}}{{1 + \sqrt 3 .1}}$ $= \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}$ $= \dfrac{{{{\left( {\sqrt 3  - 1} \right)}^2}}}{{3 - 1}}$ $= \dfrac{{4 - 2\sqrt 3 }}{2} = 2 - \sqrt 3$

Loigiaihay.com