Question

一道三角函数的题，比较难，求详解！！（高一）

Answer 1

sin^6(1°)=[sin^2(1°)]^3=[1-cos(2°)]^3/8=[1-3cos(2°)+3cos^2(2°)-cos^3(2°)]/8;
sin^6(89°)=[sin^2(89°)]^3=[1-cos(178°)]^3/8=[1+cos(2°)]^3/8
=[1+3cos(2°)+3cos^2(2°)+cos^3(2°)]/8;
所以sin^6(1°)+sin^6(89°)=[1+3cos^2(2°)]/4={1+3[cos(4°)+1]/2}/4=3cos(4°)/8+5/8
所以m/n=3cos(4°)/8+5/8+3cos(8°)/8+5/8+3cos(12°)/8+5/8+。。。+3cos(176°)/8+5/8+sin^6(45°)
=3[cos(4°)+cos(8°)+cos(12°)+...+cos(176°)]/8+[(89-1)/2]*5/8+sin^6(45°)
=3[cos(4°)+cos(176°)+cos(8°)+.cos(172°)+...+..cos(88°)+cos(92°)]/8+55/2+sin^6(45°)
=55/2+sin^6(45°)=55/2+(√2/2)^6=55/2+1/8=221/8

Answer 2

SIn89°=cos1°..........sin45°=cos45°

(sin1°)^6+sin89°^6=(sin1°)^6+(cos1°)^6=[(sin1°)^2+(cos1°)^2][sin1°^4+cos1°^4-sin1°^2*cos1°^2]
=1*[(sin1°^2+cos1°^2)^2-3sin1°^2cos1°^2]
=1-3/4*sin2°^2
同理(sin2°)^6+sin88°^6=1-3/4*sin4°^2 。。。。。。。。。。。。
∴原式= 45-3/4（sin2°^2+sin4°^2+.......sin90°^2)-sin45°^6
=45-3/4(sin2°^2+sin88°^2+sin4°^2+sin86°^2+........ +sin44°^2+sin46°^2+sin90°^2)-1/8
= 45-3/4(22+1)-1/8
=221/8
∴m/n=221/8