Video hướng dẫn giải
- LG a
- LG b
Tính
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:
\[\sin \dfrac{7\pi}{12} = \sin \left [ \dfrac{\pi}{3}+\dfrac{\pi}{4} \right ] \]
\[=\sin\dfrac{\pi }{3}\cos\dfrac{\pi}{4}+ \cos \dfrac{\pi }{3}\sin\dfrac{\pi}{4}\]
\[ = \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} + \dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2} \] \[= \dfrac{{\sqrt 6 }}{4} + \dfrac{{\sqrt 2 }}{4} = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}\]
+] \[\cos \left [ -\dfrac{\pi }{12} \right ] = \cos \left [ \dfrac{\pi }{4} -\dfrac{\pi }{3}\right ] \]
\[= \cos \dfrac{\pi }{4}\cos\dfrac{\pi }{3} + \sin \dfrac{\pi }{3}\sin \dfrac{\pi }{4}\] \[ =\dfrac{{\sqrt 2 }}{2} . \dfrac{1}{2}+ \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} \] \[= \dfrac{{\sqrt 2 }}{4} + \dfrac{{\sqrt 6 }}{4} = \dfrac{{\sqrt 2 + \sqrt 6 }}{4}\]
+] \[\tan \left [ \dfrac{13\pi }{12} \right ] = \tan[π + \dfrac{\pi }{12}] \]
\[= \tan \dfrac{\pi }{12} = \tan \left [ \dfrac{\pi }{3}-\dfrac{\pi}{4} \right ]\]
\[= \dfrac{\tan\dfrac{\pi }{3}-\tan\dfrac{\pi }{4}}{1+\tan\dfrac{\pi }{3}\tan\dfrac{\pi }{4}}\]
\[ = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 .1}} \] \[= \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \] \[= \dfrac{{{{\left[ {\sqrt 3 - 1} \right]}^2}}}{{3 - 1}} \] \[= \dfrac{{4 - 2\sqrt 3 }}{2} = 2 - \sqrt 3 \]