三角恒等变换有关数学题1.sin(π/12)-√3cos(π/12)的值?A.0 B.-√2 C.2 D.√2要详解,比如涉及那些知识点?
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![三角恒等变换有关数学题1.sin(π/12)-√3cos(π/12)的值?A.0 B.-√2 C.2 D.√2要详解,比如涉及那些知识点?](/uploads/image/z/5478247-55-7.jpg?t=%E4%B8%89%E8%A7%92%E6%81%92%E7%AD%89%E5%8F%98%E6%8D%A2%E6%9C%89%E5%85%B3%E6%95%B0%E5%AD%A6%E9%A2%981.sin%EF%BC%88%CF%80%2F12%EF%BC%89-%E2%88%9A3cos%EF%BC%88%CF%80%2F12%EF%BC%89%E7%9A%84%E5%80%BC%3FA.0+B.-%E2%88%9A2+C.2+D.%E2%88%9A2%E8%A6%81%E8%AF%A6%E8%A7%A3%2C%E6%AF%94%E5%A6%82%E6%B6%89%E5%8F%8A%E9%82%A3%E4%BA%9B%E7%9F%A5%E8%AF%86%E7%82%B9%3F)
三角恒等变换有关数学题1.sin(π/12)-√3cos(π/12)的值?A.0 B.-√2 C.2 D.√2要详解,比如涉及那些知识点?
三角恒等变换有关数学题
1.sin(π/12)-√3cos(π/12)的值?
A.0 B.-√2 C.2 D.√2
要详解,比如涉及那些知识点?
三角恒等变换有关数学题1.sin(π/12)-√3cos(π/12)的值?A.0 B.-√2 C.2 D.√2要详解,比如涉及那些知识点?
B
原式=2*(cos(π/3)*(sin(π/12)-sin(π/3)*cos(π/12))=2*sin(-π/4)=.-√2
涉及知识点:把1/2看成cos(π/3).
sin(π/12)-√3cos(π/12)
=2[1/2sin(π/12)-√3/2cos(π/12)]
=2[cos(π/3)sin(π/12)-sin(π/3)cos(π/12)]
=-2[sin(π/3)cos(π/12)-cos(π/3)sin(π/12)]
=-2sin(π/3-π/12)
=-2sin(π/4)
=-√2
选B
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sin(π/12)-√3cos(π/12)
=2[1/2sin(π/12)-√3/2cos(π/12)]
=2[cos(π/3)sin(π/12)-sin(π/3)cos(π/12)]
=-2[sin(π/3)cos(π/12)-cos(π/3)sin(π/12)]
=-2sin(π/3-π/12)
=-2sin(π/4)
=-√2
选B
其中用到了两角和与差的三角函数公式
sin(α-β)=sinαcosβ-cosαsinβ
第二种方法:
当然也可以用
cos(α+β)=cosαcosβ-sinαsinβ
只要将1/2变换成sin(π/6)
将√3/2变换成cos(π/6)
就可以了,你可以自己做一下。
(必须要自己做,才能真正的会做,祝你学习进步!)
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