| ■8700 / inTopicNo.2) |
Re[1]: 三角関数
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□投稿者/ 白拓 ファミリー(166回)-(2006/02/07(Tue) 02:04:39)
 | > sin^2/3π+sin(-7/3π)cos(-5/6π)
sin(2/3π)+sin(-7/3π)cos(-5/6π)の誤りなら、
sin(2/3π)+(1/2){sin((-7/3-5/6)π)+sin((-7/3+5/6)π)}
=sin(2/3π)+(1/2){-sin(3+1/6)π)-sin(1+1/2)π)}
=(1/2){sin(1/6)π+sin(1/2)π)}+sin(2/3π)
=3/4+√3/2
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