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A math about trigonometric

發問:

solve the following equation 7sinAcosA+cos^2(A)=2 sinA-3^0.5cosA=2^0.5 6tan^2(A)-4sin^2(A)=1

最佳解答:

As follows~ 圖片參考：http://i248.photobucket.com/albums/gg191/ncy_0916/maths88.jpg?t=1220673066 圖片參考：http://i248.photobucket.com/albums/gg191/ncy_0916/maths89.jpg?t=1220673080 2008-09-06 11:56:10 補充： a sinθ + b cosθ = r sin(θ + α) where r = √(a^2 + b^2) , tanα = b / a and 0 < α < 90* a sinθ - b cosθ = r sin(θ - α) where r = √(a^2 + b^2) , tanα = b / a and 0 < α < 90* 2008-09-06 15:10:59 補充： sorry，第1題的解是8.5度式73.4度

其他解答:

7sinAcosA+cos^2(A)=2 7sinAcosA + (cosA)^2 = 2(cosA)^2 + 2(sinA)^2 2(sinA)^2 - 7sinAcosA + (cosA)^2 = 0 sinA ={ 7cosA +/- sqrt[49(cosA)^2-8(cosA)^2] } /4 sinA = cosA { 7 +/- sqrt[49-8]}/4 tanA = {7 +/- sqrt(41) } / 4 A = tan-1 {[ 7 +/- sqrt(41) ] / 4}