高一数学中有关正弦、余弦的诱导公式已知cos(15°+α)=3/5,α为锐角,求【tan(435°-α)+sin(α-165°)】/【cos(195°+α)sin(105°+α)】的值.这是参考书上一道例题,虽然有着过程却是大致的,所以
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高一数学中有关正弦、余弦的诱导公式已知cos(15°+α)=3/5,α为锐角,求【tan(435°-α)+sin(α-165°)】/【cos(195°+α)sin(105°+α)】的值.这是参考书上一道例题,虽然有着过程却是大致的,所以
高一数学中有关正弦、余弦的诱导公式
已知cos(15°+α)=3/5,α为锐角,求【tan(435°-α)+sin(α-165°)】/【cos(195°+α)sin(105°+α)】的值.
这是参考书上一道例题,虽然有着过程却是大致的,所以我希望谁能给我详细的解答过程,并说明理由,即最后的结果,不过结果时候的计算过程就不用了.
高一数学中有关正弦、余弦的诱导公式已知cos(15°+α)=3/5,α为锐角,求【tan(435°-α)+sin(α-165°)】/【cos(195°+α)sin(105°+α)】的值.这是参考书上一道例题,虽然有着过程却是大致的,所以
【tan(435°-α)+sin(α-165°)】/【cos(195°+α)sin(105°+α)】=【tan(360°+75°-α)+sin(α-180°+15°)】/【cos(180°+15°+α)sin(90°+15°+α)】
=【tan(75°-α)-sin(α+15°)】/【cos(15°+α)cos(15°+α)】
=【tan(90°-15°-α)-sin(α+15°)】/(3/5)^2
=【cot(-15°-α)-sin(α+15°)】/(3/5)^2
=【cot(15°+α)-sin(α+15°)】/(3/5)^2
=【[sin(15°+α)-sin(α+15°)]/cos(15°+α)】/(3/5)^2
=(5/3)^3=125/27