前>>328
>>161
(1+2√(25-24cosθ)/sinθ)^2=100-96cosθ-36cos^2θ(1-cos^2θ)
1+4√(25-24cosθ)/√(1-cos^2θ)+4(25-24cosθ)/(1-cos^2θ)=100-96cosθ-36cos^2θ(1-cos^2θ)
(1-cos^2θ)+4√(25-24cosθ)√(1-cos^2θ)+4(25-24cosθ)=100(1-cos^2θ)-96cosθ(1-cos^2θ)-36cos^2θ(1-cos^2θ)^2
1-cos^2θ+4√{(25-24cosθ)(1-cos^2θ)}=-100cos^2θ+96cosθ･cos^2θ-36cos^2θ(1-cos^2θ)^2
1-cos^2θ+4√(25-24cosθ-25cos^2θ+24cos^3θ)=-100cos^2θ+96cosθ･cos^2θ-36cos^2θ(1-2cos^2θ+cos^4θ)
4√(25-24cosθ-25cos^2θ+24cos^3θ)=cos^2θ-1-100cos^2θ+96cos^3θ-36cos^2θ+72cos^4θ-36cos^6θ
4√(25-24cosθ-25cos^2θ+24cos^3θ)=-1-135cos^2θ+96cos^3θ+72cos^4θ-36cos^6θ
16(25-24cosθ-25cos^2θ+24cos^3θ)=(-1-135cos^2θ+96cos^3θ+72cos^4θ-36cos^6θ)^2
400-384cosθ-400cos^2θ+384cos^3θ=1+18225cos^4θ+9216cos^6θ+5184cos^8θ+1296cos^12θ+270cos^2θ-192cos^3θ-144cos^4θ+72cos^6θ-135･96cos^5θ-135･72cos^6θ+135･36cos^8θ+96･72cos^7θ-96･36cos^9θ-72･36cos^10θ
どうやって解くんだこれ!?　二次式じゃないら？
399-384cosθ-670cos^2θ+576cos^3θ-18081cos^4θ+12960cos^5θ+432cos^6θ-6912cos^7θ-10044cos^8θ+3456cos^9θ+2592cos^10θ-1296cos^12θ=0
cosθ≒0.0601
∴sinθ≒0.9399 うそだぁ