1、化簡求值：\(\cfrac{3-sin70^{\circ}}{2-cos^210^{\circ}}\)

分析：\(\cfrac{3-sin70^{\circ}}{2-cos^210^{\circ}}=\cfrac{3-cos20^{\circ}}{2-cos^210^{\circ}}=\cfrac{3-(2cos^210^{\circ}-1)}{2-cos^210^{\circ}}=\cfrac{2(2-cos^210^{\circ})}{2-cos^210^{\circ}}=2\)

2、化簡求值：\(4cos50^{\circ}-tan40^{\circ}\)

分析：\(4cos50^{\circ}-tan40^{\circ}=4cos50^{\circ}-\cfrac{sin40^{\circ}}{cos40^{\circ}}=\cfrac{4cos50^{\circ}cos40^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{4sin40^{\circ}cos40^{\circ}-sin40^{\circ}}{cos40^{\circ}}\)

\(=\cfrac{2sin80^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{2cos10^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{2cos(40^{\circ}-30^{\circ})-sin40^{\circ}}{cos40^{\circ}}\)

\(=\cfrac{2cos40^{\circ}\cdot \cfrac{\sqrt{3}}{2}+2sin40^{\circ}\cdot \cfrac{1}{2}-sin40^{\circ}}{cos40^{\circ}}=\sqrt{3}\).

3、化簡求值：\(\cfrac{sin8^{\circ}+sin7^{\circ}cos15^{\circ}}{cos8^{\circ}-sin7^{\circ}sin15^{\circ}}\)

分析：\(\cfrac{sin8^{\circ}+sin7^{\circ}cos15^{\circ}}{cos8^{\circ}-sin7^{\circ}sin15^{\circ}}=\cfrac{sin(15^{\circ}-7^{\circ})+sin7^{\circ}cos15^{\circ}}{cos(15^{\circ}-7^{\circ})-sin7^{\circ}sin15^{\circ}}=\cfrac{sin15^{\circ}}{cos15^{\circ}}=tan15^{\circ}=2-\sqrt{3}\).

4、化簡求值：\(cos\cfrac{\pi}{17}\cdot cos\cfrac{2\pi}{17}\cdot cos\cfrac{4\pi}{17}\cdot cos\cfrac{8\pi}{17}\)

分析：\(cos\cfrac{\pi}{17}\cdot cos\cfrac{2\pi}{17}\cdot cos\cfrac{4\pi}{17}\cdot cos\cfrac{8\pi}{17}=\cfrac{2sin\cfrac{\pi}{17}cos\cfrac{\pi}{17}\cdot cos\cfrac{2\pi}{17}\cdot cos\cfrac{4\pi}{17}\cdot cos\cfrac{8\pi}{17}}{2sin\cfrac{\pi}{17}}\)

\(=\cfrac{sin\cfrac{2\pi}{17}\cdot cos\cfrac{2\pi}{17}\cdot cos\cfrac{4\pi}{17}\cdot cos\cfrac{8\pi}{17}}{2sin\cfrac{\pi}{17}}=\cfrac{2sin\cfrac{2\pi}{17}\cdot cos\cfrac{2\pi}{17}\cdot cos\cfrac{4\pi}{17}\cdot cos\cfrac{8\pi}{17}}{2^2sin\cfrac{\pi}{17}}\)

\(=\cfrac{2sin\cfrac{4\pi}{17}\cdot cos\cfrac{4\pi}{17}\cdot cos\cfrac{8\pi}{17}}{2^3sin\cfrac{\pi}{17}}=\cfrac{2sin\cfrac{8\pi}{17}\cdot cos\cfrac{8\pi}{17}}{2^4sin\cfrac{\pi}{17}}\)

\(=\cfrac{sin\cfrac{16\pi}{17}}{2^4sin\cfrac{\pi}{17}}=\cfrac{sin\cfrac{\pi}{17}}{2^4sin\cfrac{\pi}{17}}=\cfrac{1}{16}\)

5、已知\(sinx+cosx=\cfrac{\sqrt{2}}{2}\)，化簡求值：\(sin^4x+cos^4x\)

分析：由題目可知，\((sinx+cosx)^2=(\cfrac{\sqrt{2}}{2})^2\)，即\(1+2sinxcosx=\cfrac{1}{2}\)，故\(2sinxcosx=-\cfrac{1}{2}\)

\(sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sinx^2cos^2x=1-2sinx^2cos^2x=1-\cfrac{1}{2}(2sinxcosx)^2=1-\cfrac{1}{8}=\cfrac{7}{8}\)

三角函數式的求值