作业帮求解一道三角函数题
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求解一道三角函数题
求解一道三角函数题
求解一道三角函数题
(1)sin(π-A) - cos(π+A)
=sinA + cosA
=√2sin(A+ π/4)=√2/3
∴sin(A+ π/4) = 1/3
sinA - cosA =√2sin（A-П/4)=-√2cos(A+ π/4)=4/3.
(2)sinA + cosA=√2/3
sinA - cosA =4/3.
sinA=(4+√2)/6
cos2A=1-2sin^A=-4√2/9.
∵sin(π-A) - cos(π+A)
=sinA + cosA
=(根号2)sin(A+ π/4)=(根号2)/3
∴sin(A+ π/4) = 1/3
sinA - cosA = (根号2)sin(π/4 - A) = (根号2)cos[π/2-(π/4 - A)] = (根号2)cos(π/4 + A) = 4/3