How do you evaluate sin 360°?

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Answer 1 · 2018-02-18T15:07:57

#0#

#sin 360^@#

#=sin (180*2)^@#

We know that #sin 2a=2sinacosa#

#=>2 \ sin180^@cos180^@#

but #sin180^@=0,cos180^@=-1#,

#<=>2*0*-1#

#=0#

#:.sin 360^@=0#