已知f(a)=sin(π-a)cos(2π-a)/cos(-π-a)tan(a+π)求f(-31π/3)的值
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解
f(a)=[sin(π-a)cos(2π-a)]/[cos(-π-a)tan(a+π)]
=[sinacos(-a)]/[cos(π+a)tana]
=[sinacosa]/[-cosatana]
=[sinacosa]/(-sina)
=-cosa
f(-31π/3)
=-cos(-31π/3)
=-cos(31π/3)
=-cos(10π+π/3)
=-cos(π/3)
=-1/2
f(a)=[sin(π-a)cos(2π-a)]/[cos(-π-a)tan(a+π)]
=[sinacos(-a)]/[cos(π+a)tana]
=[sinacosa]/[-cosatana]
=[sinacosa]/(-sina)
=-cosa
f(-31π/3)
=-cos(-31π/3)
=-cos(31π/3)
=-cos(10π+π/3)
=-cos(π/3)
=-1/2
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