Question

Вычислите: 1) 16(cos^2x-sin^2x), если x=π/6 2) 8(sin^2x-cos^2x), если x=π/6

Answer 1

$6(cos^2x-sin^2x)=6*cos(2x)=6*cos(2* \frac{\pi}{6} )=\\= 6*cos(60^0)=6* \frac{1}{2} =3$
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$8(sin^2x-cos^2x)=-8(cos^2x-sin^2x)=-8*cos(2x)=\\= -8*cos(60^0)=-4$