Question

高一数学，一道三角函数求值的题，要过程！

Answer 1

原式=sin(π/6)+cos(π/3)+tan(-π/4)
=1/2+1/2-1
=0

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Answer 2

首先要知道 sin(α+2kπ)=sinα ，cos(α+2kπ)=cosα，tan(α+kπ)=tanα，tan(-α) = -tanα
所以sin(25π/6)=sin(25π/6 - 4π)=sin(π/6) =1/2
cos(25π/3)=cos(25π/3-8π)=cos(π/3)=1/2
tan(-25π/4)=tan(-25π/4+6π)=tan(-π/4)=-tan(π/4)=-1
所以原始=1/2+1/2-1=0

Answer 3

sin(25π/6)+cos(25π/3)+tan(-25π/4)
=sin(4π+π/6)+cos(8π+π/3)+tan(-6π-π/4)
=sin(π/6)+cos(π/3)+tan(-π/4)=sin(π/6)+cos(π/3)-tan(π/4)
=1/2+1/2-1
=0

Answer 4

sin(25π/6)+cos(25π/3)+tan(-25π/4)
=sin(π/6)+cos(π/3)+tan(-π/4)
=sin(π/6)+cos(π/3)-tan(π/4)
=1/2+1/2-1
=0

Answer 5

=sin(4π+π/6)+cos(8π+π/3)+tan(-6π-π/4)
=sin(π/6)+cos(π/3)+tan(-π/4)
=1/2+1/2-1=0