解答
tan(3y+60)tan(2y+5)=1
解答
y=−13+52πn+10π,y=−13+52πn+103π
+1
度数
y=−726.84513…∘+72∘n,y=−690.84513…∘+72∘n求解步骤
tan(3y+60)tan(2y+5)=1
两边减去 1tan(3y+60)tan(2y+5)−1=0
用 sin, cos 表示
−1+tan(5+2y)tan(60+3y)
使用基本三角恒等式: tan(x)=cos(x)sin(x)=−1+cos(5+2y)sin(5+2y)tan(60+3y)
使用基本三角恒等式: tan(x)=cos(x)sin(x)=−1+cos(5+2y)sin(5+2y)⋅cos(60+3y)sin(60+3y)
化简 −1+cos(5+2y)sin(5+2y)⋅cos(60+3y)sin(60+3y):cos(5+2y)cos(60+3y)−cos(5+2y)cos(60+3y)+sin(5+2y)sin(60+3y)
−1+cos(5+2y)sin(5+2y)⋅cos(60+3y)sin(60+3y)
乘 cos(5+2y)sin(5+2y)⋅cos(60+3y)sin(60+3y):cos(2y+5)cos(3y+60)sin(2y+5)sin(3y+60)
cos(5+2y)sin(5+2y)⋅cos(60+3y)sin(60+3y)
分式相乘: ba⋅dc=b⋅da⋅c=cos(5+2y)cos(60+3y)sin(5+2y)sin(60+3y)
=−1+cos(2y+5)cos(3y+60)sin(2y+5)sin(3y+60)
将项转换为分式: 1=cos(5+2y)cos(60+3y)1cos(5+2y)cos(60+3y)=−cos(5+2y)cos(60+3y)1⋅cos(5+2y)cos(60+3y)+cos(5+2y)cos(60+3y)sin(5+2y)sin(60+3y)
因为分母相等,所以合并分式: ca±cb=ca±b=cos(5+2y)cos(60+3y)−1⋅cos(5+2y)cos(60+3y)+sin(5+2y)sin(60+3y)
乘以:1⋅cos(5+2y)=cos(5+2y)=cos(2y+5)cos(3y+60)−cos(2y+5)cos(3y+60)+sin(2y+5)sin(3y+60)
=cos(5+2y)cos(60+3y)−cos(5+2y)cos(60+3y)+sin(5+2y)sin(60+3y)
cos(5+2y)cos(60+3y)−cos(5+2y)cos(60+3y)+sin(5+2y)sin(60+3y)=0
g(x)f(x)=0⇒f(x)=0−cos(5+2y)cos(60+3y)+sin(5+2y)sin(60+3y)=0
使用三角恒等式改写
−cos(5+2y)cos(60+3y)+sin(5+2y)sin(60+3y)
使用角和恒等式: cos(s)cos(t)−sin(s)sin(t)=cos(s+t)−cos(s)cos(t)+sin(s)sin(t)=−cos(s+t)=−cos(5+2y+60+3y)
−cos(5+2y+60+3y)=0
两边除以 −1
−cos(5+2y+60+3y)=0
两边除以 −1−1−cos(5+2y+60+3y)=−10
化简cos(5+2y+60+3y)=0
cos(5+2y+60+3y)=0
cos(5+2y+60+3y)=0的通解
cos(x) 周期表(周期为 2πn):
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
5+2y+60+3y=2π+2πn,5+2y+60+3y=23π+2πn
5+2y+60+3y=2π+2πn,5+2y+60+3y=23π+2πn
解 5+2y+60+3y=2π+2πn:y=−13+52πn+10π
5+2y+60+3y=2π+2πn
对同类项分组2y+3y+5+60=2π+2πn
同类项相加:2y+3y=5y5y+5+60=2π+2πn
数字相加:5+60=655y+65=2π+2πn
将 65到右边
5y+65=2π+2πn
两边减去 655y+65−65=2π+2πn−65
化简5y=2π+2πn−65
5y=2π+2πn−65
两边除以 5
5y=2π+2πn−65
两边除以 555y=52π+52πn−565
化简
55y=52π+52πn−565
化简 55y:y
55y
数字相除:55=1=y
化简 52π+52πn−565:−13+52πn+10π
52π+52πn−565
对同类项分组=−565+52πn+52π
565=13
565
数字相除:565=13=13
52π=10π
52π
使用分式法则: acb=c⋅ab=2⋅5π
数字相乘:2⋅5=10=10π
=−13+52πn+10π
y=−13+52πn+10π
y=−13+52πn+10π
y=−13+52πn+10π
解 5+2y+60+3y=23π+2πn:y=−13+52πn+103π
5+2y+60+3y=23π+2πn
对同类项分组2y+3y+5+60=23π+2πn
同类项相加:2y+3y=5y5y+5+60=23π+2πn
数字相加:5+60=655y+65=23π+2πn
将 65到右边
5y+65=23π+2πn
两边减去 655y+65−65=23π+2πn−65
化简5y=23π+2πn−65
5y=23π+2πn−65
两边除以 5
5y=23π+2πn−65
两边除以 555y=523π+52πn−565
化简
55y=523π+52πn−565
化简 55y:y
55y
数字相除:55=1=y
化简 523π+52πn−565:−13+52πn+103π
523π+52πn−565
对同类项分组=−565+52πn+523π
565=13
565
数字相除:565=13=13
523π=103π
523π
使用分式法则: acb=c⋅ab=2⋅53π
数字相乘:2⋅5=10=103π
=−13+52πn+103π
y=−13+52πn+103π
y=−13+52πn+103π
y=−13+52πn+103π
y=−13+52πn+10π,y=−13+52πn+103π