✨ Best Answer ✨
f(θ)=2sin(θ+π/6)
g(θ)=-2cos(2θ+π/3)
cos2a=1-2sin^2a
2a=2θ+π/3のとき
a=θ+π/6
つまりcos(2θ+π/3)=1-2sin^2(θ+π/6)
g(θ)=-2{1-2sin^2(θ+π/6)}
=4sin^2(θ+π/6)-2
よってg(θ)=2{f(θ)}^2-2
訂正
g(θ)={f(θ)}^2-2です
ありがとうございます!
✨ Best Answer ✨
f(θ)=2sin(θ+π/6)
g(θ)=-2cos(2θ+π/3)
cos2a=1-2sin^2a
2a=2θ+π/3のとき
a=θ+π/6
つまりcos(2θ+π/3)=1-2sin^2(θ+π/6)
g(θ)=-2{1-2sin^2(θ+π/6)}
=4sin^2(θ+π/6)-2
よってg(θ)=2{f(θ)}^2-2
訂正
g(θ)={f(θ)}^2-2です
ありがとうございます!
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ありがとうございます!