利用和(差)角公式求下列各三角函数的值 (1) sin(-7π/12) (2)cos(-61π/12) (3)tan(3
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利用和(差)角公式求下列各三角函数的值 (1) sin(-7π/12) (2)cos(-61π/12) (3)tan(35π/12
![利用和(差)角公式求下列各三角函数的值 (1) sin(-7π/12) (2)cos(-61π/12) (3)tan(3](/uploads/image/z/149780-20-0.jpg?t=%E5%88%A9%E7%94%A8%E5%92%8C%EF%BC%88%E5%B7%AE%EF%BC%89%E8%A7%92%E5%85%AC%E5%BC%8F%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E7%9A%84%E5%80%BC+%EF%BC%881%EF%BC%89+sin%EF%BC%88-7%CF%80%2F12%EF%BC%89+%EF%BC%882%EF%BC%89cos%EF%BC%88-61%CF%80%2F12%EF%BC%89+%EF%BC%883%EF%BC%89tan%EF%BC%883)
sin(-7π/12)
= -sin(7π/12)
= -sin﹙π/3+π/4﹚
= -(sinπ/3cosπ/4+cosπ/3sinπ/4)
= -(√3/2 * √2/2 + 1/2 * √2/2)
= -(√6+√2)/4
cos(-61π/12)
= cos(61π/12)
= cos(5π+π/12)
= cos(π+π/12)
= -cos(π/12)
= -cos(π/3-π/4)
= -(cosπ/3cosπ/4+sinπ/3sinπ/4)
= -(1/2 * √2/2 + √3/2 * √2/2)
= -(√2+√6)/4
tan(35π/12)
= tan(3π-π/12)
= tan(π-π/12)
= -tan(π/12)
= -tan(π/3-π/4)
= -(tanπ/3-tanπ/4) / (1+tanπ/3tanπ/4)
= -(√3-1) / (1+√3)
= -(√3-1)^2 /{ (√3+1)(√3-1)}
= -(4-2√3)/(3-1)
= √3 - 2
= -sin(7π/12)
= -sin﹙π/3+π/4﹚
= -(sinπ/3cosπ/4+cosπ/3sinπ/4)
= -(√3/2 * √2/2 + 1/2 * √2/2)
= -(√6+√2)/4
cos(-61π/12)
= cos(61π/12)
= cos(5π+π/12)
= cos(π+π/12)
= -cos(π/12)
= -cos(π/3-π/4)
= -(cosπ/3cosπ/4+sinπ/3sinπ/4)
= -(1/2 * √2/2 + √3/2 * √2/2)
= -(√2+√6)/4
tan(35π/12)
= tan(3π-π/12)
= tan(π-π/12)
= -tan(π/12)
= -tan(π/3-π/4)
= -(tanπ/3-tanπ/4) / (1+tanπ/3tanπ/4)
= -(√3-1) / (1+√3)
= -(√3-1)^2 /{ (√3+1)(√3-1)}
= -(4-2√3)/(3-1)
= √3 - 2
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