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)#