解答
196sin(θ)−49cos(θ)=160
解答
θ=2.47257…+2πn,θ=1.15897…+2πn
+1
度数
θ=141.66783…∘+360∘n,θ=66.40464…∘+360∘n求解步骤
196sin(θ)−49cos(θ)=160
两边加上 49cos(θ)196sin(θ)=160+49cos(θ)
两边进行平方(196sin(θ))2=(160+49cos(θ))2
两边减去 (160+49cos(θ))238416sin2(θ)−25600−15680cos(θ)−2401cos2(θ)=0
使用三角恒等式改写
−25600−15680cos(θ)−2401cos2(θ)+38416sin2(θ)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−25600−15680cos(θ)−2401cos2(θ)+38416(1−cos2(θ))
化简 −25600−15680cos(θ)−2401cos2(θ)+38416(1−cos2(θ)):−40817cos2(θ)−15680cos(θ)+12816
−25600−15680cos(θ)−2401cos2(θ)+38416(1−cos2(θ))
乘开 38416(1−cos2(θ)):38416−38416cos2(θ)
38416(1−cos2(θ))
使用分配律: a(b−c)=ab−aca=38416,b=1,c=cos2(θ)=38416⋅1−38416cos2(θ)
数字相乘:38416⋅1=38416=38416−38416cos2(θ)
=−25600−15680cos(θ)−2401cos2(θ)+38416−38416cos2(θ)
化简 −25600−15680cos(θ)−2401cos2(θ)+38416−38416cos2(θ):−40817cos2(θ)−15680cos(θ)+12816
−25600−15680cos(θ)−2401cos2(θ)+38416−38416cos2(θ)
对同类项分组=−15680cos(θ)−2401cos2(θ)−38416cos2(θ)−25600+38416
同类项相加:−2401cos2(θ)−38416cos2(θ)=−40817cos2(θ)=−15680cos(θ)−40817cos2(θ)−25600+38416
数字相加/相减:−25600+38416=12816=−40817cos2(θ)−15680cos(θ)+12816
=−40817cos2(θ)−15680cos(θ)+12816
=−40817cos2(θ)−15680cos(θ)+12816
12816−15680cos(θ)−40817cos2(θ)=0
用替代法求解
12816−15680cos(θ)−40817cos2(θ)=0
令:cos(θ)=u12816−15680u−40817u2=0
12816−15680u−40817u2=0:u=−8163415680+2338305088,u=816342338305088−15680
12816−15680u−40817u2=0
改写成标准形式 ax2+bx+c=0−40817u2−15680u+12816=0
使用求根公式求解
−40817u2−15680u+12816=0
二次方程求根公式:
若 a=−40817,b=−15680,c=12816u1,2=2(−40817)−(−15680)±(−15680)2−4(−40817)⋅12816
u1,2=2(−40817)−(−15680)±(−15680)2−4(−40817)⋅12816
(−15680)2−4(−40817)⋅12816=2338305088
(−15680)2−4(−40817)⋅12816
使用法则 −(−a)=a=(−15680)2+4⋅40817⋅12816
使用指数法则: (−a)n=an,若 n 是偶数(−15680)2=156802=156802+4⋅40817⋅12816
数字相乘:4⋅40817⋅12816=2092442688=156802+2092442688
156802=245862400=245862400+2092442688
数字相加:245862400+2092442688=2338305088=2338305088
u1,2=2(−40817)−(−15680)±2338305088
将解分隔开u1=2(−40817)−(−15680)+2338305088,u2=2(−40817)−(−15680)−2338305088
u=2(−40817)−(−15680)+2338305088:−8163415680+2338305088
2(−40817)−(−15680)+2338305088
去除括号: (−a)=−a,−(−a)=a=−2⋅4081715680+2338305088
数字相乘:2⋅40817=81634=−8163415680+2338305088
使用分式法则: −ba=−ba=−8163415680+2338305088
u=2(−40817)−(−15680)−2338305088:816342338305088−15680
2(−40817)−(−15680)−2338305088
去除括号: (−a)=−a,−(−a)=a=−2⋅4081715680−2338305088
数字相乘:2⋅40817=81634=−8163415680−2338305088
使用分式法则: −b−a=ba15680−2338305088=−(2338305088−15680)=816342338305088−15680
二次方程组的解是:u=−8163415680+2338305088,u=816342338305088−15680
u=cos(θ)代回cos(θ)=−8163415680+2338305088,cos(θ)=816342338305088−15680
cos(θ)=−8163415680+2338305088,cos(θ)=816342338305088−15680
cos(θ)=−8163415680+2338305088:θ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn
cos(θ)=−8163415680+2338305088
使用反三角函数性质
cos(θ)=−8163415680+2338305088
cos(θ)=−8163415680+2338305088的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn
θ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn
cos(θ)=816342338305088−15680:θ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
cos(θ)=816342338305088−15680
使用反三角函数性质
cos(θ)=816342338305088−15680
cos(θ)=816342338305088−15680的通解cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
θ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
合并所有解θ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn,θ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
将解代入原方程进行验证
将它们代入 196sin(θ)−49cos(θ)=160检验解是否符合
去除与方程不符的解。
检验 arccos(−8163415680+2338305088)+2πn的解:真
arccos(−8163415680+2338305088)+2πn
代入 n=1arccos(−8163415680+2338305088)+2π1
对于 196sin(θ)−49cos(θ)=160代入θ=arccos(−8163415680+2338305088)+2π1196sin(arccos(−8163415680+2338305088)+2π1)−49cos(arccos(−8163415680+2338305088)+2π1)=160
整理后得160=160
⇒真
检验 −arccos(−8163415680+2338305088)+2πn的解:假
−arccos(−8163415680+2338305088)+2πn
代入 n=1−arccos(−8163415680+2338305088)+2π1
对于 196sin(θ)−49cos(θ)=160代入θ=−arccos(−8163415680+2338305088)+2π1196sin(−arccos(−8163415680+2338305088)+2π1)−49cos(−arccos(−8163415680+2338305088)+2π1)=160
整理后得−83.12602…=160
⇒假
检验 arccos(816342338305088−15680)+2πn的解:真
arccos(816342338305088−15680)+2πn
代入 n=1arccos(816342338305088−15680)+2π1
对于 196sin(θ)−49cos(θ)=160代入θ=arccos(816342338305088−15680)+2π1196sin(arccos(816342338305088−15680)+2π1)−49cos(arccos(816342338305088−15680)+2π1)=160
整理后得160=160
⇒真
检验 2π−arccos(816342338305088−15680)+2πn的解:假
2π−arccos(816342338305088−15680)+2πn
代入 n=12π−arccos(816342338305088−15680)+2π1
对于 196sin(θ)−49cos(θ)=160代入θ=2π−arccos(816342338305088−15680)+2π1196sin(2π−arccos(816342338305088−15680)+2π1)−49cos(2π−arccos(816342338305088−15680)+2π1)=160
整理后得−199.22691…=160
⇒假
θ=arccos(−8163415680+2338305088)+2πn,θ=arccos(816342338305088−15680)+2πn
以小数形式表示解θ=2.47257…+2πn,θ=1.15897…+2πn