How do you determine #sin 345#?

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Answer 1 · 2018-04-21T19:16:58

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we know that #345=300+45#

Now let use the formula

#sin(a+b)=sinacosb+sinbcosa# in our case

#sin(300+45)=sin300cos45+sin45cos300#

But #sin300=-sin60=-sqrt3/2# and #cos300=cos60=1/2# and #sin45=cos45=sqrt2/2#

Lets go to prior formula

#sin345=sin(300+45)=#
#=-sqrt3/2·sqrt2/2+sqrt2/2·1/2=sqrt2/4-sqrt6/4=1/4(sqrt2-sqrt6)#