简单三角函数题,求解
已知cosθ=3/5,θ∈(0,π/2) (1)求sin(θ+π/4) (2)求sin(2θ+π/2)
解:θ∈(0,π/2) cosθ=3/5,sinθ=4/5 (1)sin(θ+π/4)=sinθcosπ/4+cosθsinπ/4) =4/5×√2/2+3/5×√2/2 =7√2/10。 sin(2θ+π/2) =sin2θ×cosπ/2+cos2θ×sinπ/2 =sin2θ×0+(2cos²θ-1)×1 =0+[2×(3/5)²-1] =-7/25。
cosθ=3/5,θ∈(0,π/2)则sinθ为正值4/5 sin(θ+π/4)=sinθ*cosπ/4+sinπ/4*cosθ=7√2/10 cos2θ=-1/5 sinπ/2=1 cosπ/2=0 sin(2θ+π/2)=sinπ/2*cosθ=-1/5
∵θ∈(0,π/2),cosθ=3/5 ∴sinθ=4/5 ∵sin(90°+X)=cosX ①sin(θ+π/4)=(√2/2)(sinθ+cosθ)=7√2/10 ②sin(2θ+π/2)=cos2θ=2cos²θ-1=-7/25
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