How do you find the exact value of $cos(120^circ)+cos(45^circ)$ ?

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Answer 1 · 2017-03-02T09:24:17

$=-1/2(1-sqrt2)$

$cos120^0+cos45^0$

this can be deduced from the standard angles for $"cosine"$

$cos(180-theta)=-costheta$

$:.cos120^0=-cos60^0=-1/2$

$cos45^0=1/sqrt2=sqrt2/2$

$cos120^0+cos45^0=-1/2+sqrt2/2$

$=-1/2(1-sqrt2)$

How do you find the exact value of cos(120^circ)+cos(45^circ)cos(120^circ)+cos(45^circ)?