Giải bài 6 trang 156 – SGK môn Đại số lớp 10

Không sử dụng máy tính, hãy chứng minh
a) \(\sin {{75}^{o}}+\cos {{75}^{o}}=\dfrac{\sqrt{6}}{2};\)
b) \(\tan {{267}^{o}}+\tan {{93}^{o}}=0;\)
c) \(\sin {{65}^{o}}+\sin {{55}^{o}}=\sqrt{3}\cos{{5}^{o}};\)
d) \(\cos {{12}^{o}}-\cos {{48}^{o}}=\sin {{18}^{o}}.\)
Lời giải:
Gợi ý:
Biến đổi về các cung có liên quan đặc biệt.
a)
\(\sin {{75}^{o}}+\cos {{75}^{o}}=\sqrt{2}\sin \left( {{45}^{o}}+{{75}^{o}} \right)=\sqrt{2}\sin {{120}^{o}}\\=\sqrt{2}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{6}}{2}\)
\( \begin{aligned} b)&\,\tan {{267}^{o}}+\tan {{93}^{o}}\\&=\tan \left( {{180}^{o}}+{{87}^{o}} \right)+\tan \left( {{180}^{o}}-{{87}^{o}} \right) \\ & =\tan {{87}^{o}}-\tan {{87}^{o}}=0 \\ \end{aligned} \)
\( \begin{aligned} c)&\,\sin {{65}^{o}}+\sin {{55}^{o}}=\sin {{65}^{o}}+\sin \left( {{120}^{o}}-{{65}^{o}} \right) \\ & =\sin {{65}^{o}}+\dfrac{\sqrt{3}}{2}\cos {{65}^{o}}+\dfrac{1}{2}\sin {{65}^{o}} \\ & =\dfrac{3}{2}\sin {{65}^{o}}+\dfrac{\sqrt{3}}{2}\cos {{65}^{o}} \\ & =\sqrt{3}\left( \dfrac{\sqrt{3}}{2}\sin {{65}^{o}}+\dfrac{1}{2}\cos {{65}^{o}} \right) \\ & =\sqrt{3}\cos \left( {{65}^{o}}-{{60}^{o}} \right) \\ & =\sqrt{3}\cos{{5}^{o}} \\ \end{aligned} \)
\( \begin{aligned} d)&\,\cos {{12}^{o}}-\cos {{48}^{o}}=\cos {{12}^{o}}-\cos \left( {{60}^{o}}-{{12}^{o}} \right) \\ & =\cos {{12}^{o}}-\dfrac{1}{2}\cos {{12}^{o}}-\dfrac{\sqrt{3}}{2}\sin {{12}^{o}} \\ & =\dfrac{1}{2}\cos {{12}^{o}}-\dfrac{\sqrt{3}}{2}\sin {{12}^{o}} \\ & =\sin \left( {{30}^{o}}-{{12}^{o}} \right) \\ & =\sin {{18}^{o}} \\ \end{aligned} \)