Question

若sinθ+cosθ=根号2,则tan(θ+π/3)的值是 要过程

Answer 1

注用a代替θ,可得：
sina+cosa=√2 可得：
√2(√2/2sina+√2/2cosa)=√2

√2sin(a+π/4)=√2
即：sin(a+π/4)=1
所以可得：a+π/4=π/2 解得：a=π/4
tan(a+π/3)
=(tana+tanπ/3)/(1-tanatanπ/3)
=(1+√3)/(1-√3)
=-2-√3

Answer 2

由已知得 √2*(sinθ*√2/2+cosθ*√2/2)=√2 ，
因此 sin(θ+π/4)=1 ，
所以 θ=π/4+2kπ ，则 sinθ=cosθ=√2/2 ，
那么 tanθ=sinθ/cosθ=1 ，
所以 tan(θ+π/3)=[tanθ+tan(π/3)]/[1-tanθtan(π/3)]=(1+√3)/(1-√3)= -2-√3 。

Answer 3

解：∵sinθ+cosθ=√2
∴√2[（sinθ√2/2)+(√2/2)cosθ]=√2
sin(θ+π/4)=1
∴θ=π/4
∴tan(θ+π/3)=tan(π/4+π/3)
=sin(π/4+π/3)/cos(π/4+π/3)
=(1+√3)/(1-√3)
=-2-√3

Answer 4

sinθ*√2 /2+cosθ*√2 /2=1
sinθ*cos(π/4)+cosθ*sin(π/4)=1
sin(θ+π/4)=1
θ+π/4=(2k+1/2)π (K∈整数）
θ=(2k+1/4)π (K∈整数）
tanθ=1
tan(θ+π/3)=[tanθ+tan(π/3)]/[1-tanθ*tan(π/3)]
=-(2+√3)

Answer 5

sinθ+cosθ=根号2
根号2sin(θ+π/4)=根号2
sin(θ+π/4)=1
θ+π/4=2Kπ+π/2
θ=2Kπ+π/4
tan(θ+π/3)=tan7π/12=tan75=(tan30+tan45)/(1-tan30tan45)=2+根号3