請問聯立三角函數是否能手算解?

[eguitar]

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▼ 參考資料

聯立三角函數
3=cos(A)+2.3094cos(B)
sin(A)=2.3094SIN(B)

解A B 是否能手算解?

謝謝

[紫煌]

聯立三角函數
3=cos(A)+2.3094cos(B)
sin(A)=2.3094SIN(B)

解A B 是否能手算解?

謝謝

3=cos(A)+2.3094cos(B)
`(3-2.3094*cos(B))^2=1-(sin(A))^2`
`(3-2.3094*cos(B))^2=1-(2.3094*SIN(B))^2`
`(3-2.3094*cos(B))^2=1-(2.3094)^2*(1-(cos(B))^2)`

[eguitar]

聯立三角函數
3=cos(A)+2.3094cos(B)
sin(A)=2.3094SIN(B)

解A B 是否能手算解?

謝謝

3=cos(A)+2.3094cos(B)
`(3-2.3094*cos(B))^2=1-(sin(A))^2`
`(3-2.3094*cos(B))^2=1-(2.3094*SIN(B))^2`
`(3-2.3094*cos(B))^2=1-(2.3094)^2*(1-(cos(B))^2)`

原來如此 感謝

[laimaddux31]

聯立三角函數
3=cos(A)+2.3094cos(B)----->(1
sin(A)=2.3094SIN(B)----->(2

解A B 是否能手算解?

謝謝

(一)(1)式sin(A)=2.3094SIN(B)
=>`cosA=sqrt(1-2.3094^2*sinB^2)`----->(3
(二)(3)式代入(1)式
得3=`sqrt(1-2.3094^2*sinB^2)`+2.3094cos(B)
移項得3-2.3094cos(B)=`sqrt(1-2.3094^2*sinB^2)`
各開方得`(3-2.3094cos(B))^2=1-2.3094^2*sinB^2`
移項得cosB=`(1-2.3094^2-9)/(-2X3X2.3094)=0.962`
`cos^(-1)(0.962)=B=15.793`度
B代回(2)式可得A

以上的過程也可以...參考看看吧