sin(a+π/3)=-3/5sin(2a+π/6)
展开全部
我们可以利用三角函数的和角公式和倍角公式来解这个方程。
首先,根据正弦函数的和角公式,有:
sin(a + π/3) = sin(a)cos(π/3) + cos(a)sin(π/3) = (1/2)sin(a) + (sqrt(3)/2)cos(a)
其次,根据正弦函数的倍角公式,有:
sin(2a + π/6) = 2sin(a + π/12)cos(a - π/12)
将以上两个式子代入原方程,得到:
(1/2)sin(a) + (sqrt(3)/2)cos(a) = -(3/5) * 2sin(a + π/12)cos(a - π/12)
化简可得:
5sin(a) + 5sqrt(3)cos(a) = -6sin(a + π/12)cos(a - π/12)
利用正弦函数的差角公式,有:
cos(a - π/12) = cos(a)cos(π/12) + sin(a)sin(π/12) = (sqrt(6)+sqrt(2))/4*cos(a) + (sqrt(6)-sqrt(2))/4*sin(a)
sin(a + π/12) = sin(a)cos(π/12) + cos(a)sin(π/12) = (sqrt(6)-sqrt(2))/4*cos(a) + (sqrt(6)+sqrt(2))/4*sin(a)
将上式代入原方程,得到:
5sin(a) + 5sqrt(3)cos(a) = -6[(sqrt(6)-sqrt(2))/4*cos(a) + (sqrt(6)+sqrt(2))/4*sin(a)][(sqrt(6)+sqrt(2))/4*cos(a) - (sqrt(6)-sqrt(2))/4*sin(a)]
化简可得:
25sin(a) + 15sqrt(3)cos(a) = -9sqrt(2)sin(a) - 9sqrt(6)cos(a)
移项并整理可得:
(9sqrt(6) + 25)sin(a) = (-15sqrt(3) - 9sqrt(2))cos(a)
将两边同时除以 cos(a),并根据三角函数的平方和公式,可得:
tan(a) = -3/5
因此,a 的值可以取反正切函数的值:
a = atan(-3/5)
最后,将 a 的值代入原方程中的任意一个式子中,即可求出 sin(a+π/3) 的值,进而求解出这个方程的解。
首先,根据正弦函数的和角公式,有:
sin(a + π/3) = sin(a)cos(π/3) + cos(a)sin(π/3) = (1/2)sin(a) + (sqrt(3)/2)cos(a)
其次,根据正弦函数的倍角公式,有:
sin(2a + π/6) = 2sin(a + π/12)cos(a - π/12)
将以上两个式子代入原方程,得到:
(1/2)sin(a) + (sqrt(3)/2)cos(a) = -(3/5) * 2sin(a + π/12)cos(a - π/12)
化简可得:
5sin(a) + 5sqrt(3)cos(a) = -6sin(a + π/12)cos(a - π/12)
利用正弦函数的差角公式,有:
cos(a - π/12) = cos(a)cos(π/12) + sin(a)sin(π/12) = (sqrt(6)+sqrt(2))/4*cos(a) + (sqrt(6)-sqrt(2))/4*sin(a)
sin(a + π/12) = sin(a)cos(π/12) + cos(a)sin(π/12) = (sqrt(6)-sqrt(2))/4*cos(a) + (sqrt(6)+sqrt(2))/4*sin(a)
将上式代入原方程,得到:
5sin(a) + 5sqrt(3)cos(a) = -6[(sqrt(6)-sqrt(2))/4*cos(a) + (sqrt(6)+sqrt(2))/4*sin(a)][(sqrt(6)+sqrt(2))/4*cos(a) - (sqrt(6)-sqrt(2))/4*sin(a)]
化简可得:
25sin(a) + 15sqrt(3)cos(a) = -9sqrt(2)sin(a) - 9sqrt(6)cos(a)
移项并整理可得:
(9sqrt(6) + 25)sin(a) = (-15sqrt(3) - 9sqrt(2))cos(a)
将两边同时除以 cos(a),并根据三角函数的平方和公式,可得:
tan(a) = -3/5
因此,a 的值可以取反正切函数的值:
a = atan(-3/5)
最后,将 a 的值代入原方程中的任意一个式子中,即可求出 sin(a+π/3) 的值,进而求解出这个方程的解。
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询