Cos80°-cos40°+√3cos70°=?

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Answer 1 · 2018-04-21T20:59:16

$cos80^@-cos40^@+sqrt3cos70^@=0$

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$cos80^@-cos40^@+sqrt3cos70^@=?$

$cosA-cosB=-2sin((A+B)/2)sin((A-B)/2)$

$cos80^@-cos40^@=-2sin((80^@+40^@)/2)sin((80^@-40^@)/2)=-2sin60^@sin20^@=-2(sqrt3)/2sin20^@=-sqrt3sin20^@$

Let's plug this in:

$cos80^@-cos40^@+sqrt3cos70^@=-sqrt3sin20^@+sqrt3cos70^@=sqrt3(sin20^@-cos70^@)$

$costheta=sin(pi/2-theta)$

$cos70^@=sin(90^@-70^@)=sin20^@$

Let's plug this in:

$cos80^@-cos40^@+sqrt3cos70^@=sqrt3(sin20^@-cos70^@)=sqrt3(sin20^@-sin20^@)=0$