1.With their bare hands,they fought to save the man who was seriously hurt in the attack.2.三角函数:sin(60°-x)cos(x-20°)+cos(65°-x)cos(110°-x)的值为( )A.√2 B.(√2)/2 C.1/2 D.(√3)/2请写出具体过程
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![1.With their bare hands,they fought to save the man who was seriously hurt in the attack.2.三角函数:sin(60°-x)cos(x-20°)+cos(65°-x)cos(110°-x)的值为( )A.√2 B.(√2)/2 C.1/2 D.(√3)/2请写出具体过程](/uploads/image/z/11367990-54-0.jpg?t=1.With+their+bare+hands%2Cthey+fought+to+save+the+man+who+was+seriously+hurt+in+the+attack.2.%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%3Asin%2860%C2%B0-x%29cos%28x-20%C2%B0%29%2Bcos%2865%C2%B0-x%29cos%28110%C2%B0-x%29%E7%9A%84%E5%80%BC%E4%B8%BA%28+%29A.%E2%88%9A2+B.%28%E2%88%9A2%29%EF%BC%8F2+C.1%EF%BC%8F2+D.%28%E2%88%9A3%29%EF%BC%8F2%E8%AF%B7%E5%86%99%E5%87%BA%E5%85%B7%E4%BD%93%E8%BF%87%E7%A8%8B)
1.With their bare hands,they fought to save the man who was seriously hurt in the attack.2.三角函数:sin(60°-x)cos(x-20°)+cos(65°-x)cos(110°-x)的值为( )A.√2 B.(√2)/2 C.1/2 D.(√3)/2请写出具体过程
1.
With their bare hands,they fought to save the man who was seriously hurt in the attack.
2.三角函数:
sin(60°-x)cos(x-20°)+cos(65°-x)cos(110°-x)的值为( )
A.√2 B.(√2)/2 C.1/2 D.(√3)/2
请写出具体过程
1.With their bare hands,they fought to save the man who was seriously hurt in the attack.2.三角函数:sin(60°-x)cos(x-20°)+cos(65°-x)cos(110°-x)的值为( )A.√2 B.(√2)/2 C.1/2 D.(√3)/2请写出具体过程
1.他们赤手空拳地拼着命,就是为了挽救那个在战斗中严重受伤的男人.
2.sin(65°-x)cos(x-20°)+cos(65°-x)cos(110°-x)=sin(65°-x)cos(x-20°)+cos(65°-x)sin(x-20°)
=sin[(65°-x)+(x-20°)]=sin45°=根号2 /2
问题不是很简单的话, 一般有悬赏比较好。
他们赤手空拳拼命挽救那个在战斗中严重受伤的男人。
第二道题题目可能写错了 前面那部分应该都是sin
这样结果就是B。
用他们赤裸的双手,他们奋勇抢救在袭击中受伤很严重的男子
他们赤手空拳地,救出了在战争中严重受伤的男人。
B