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

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

How do you determine sin 345sin 345?