數學問題 10分

1)畫個直角三角形

sinθ = sqrt(37^2-12^2)/37 = 32/37 (sqrt=根號)
tanθ = 35/12

[5 / sinθ] + 4tanθ = 5/(32/37)+4(35/12) = 356/21

2)
[1 / 1+tan^2θ] + sin^2θ
=[ 1 / 1+(sin^2θ/cos^2θ) ] + sin^2θ
=[ cos^2θ / (cos^2θ + sin^2θ) ] +sin^2θ
=cos^2θ/1 + sin^2θ
=1

3) cos64° = cos(90°-26°) = sin26°
∴ sin(θ - 10°) = sin26°
θ - 10°=26°
θ =36°

1. Let x be the length of opposite side of the right-angled triangle.
By Pyth. theorem,
x^2 = 37^2 - 12^2
x=35
therefore, sinθ = 35/37 , tanθ = 35/12
Hence, [5 / sinθ] + 4tanθ = [5 / (35/37)] + 4(35/12) = 356/21

2. .[1 / 1+tan^2 θ]+sin^2 θ = (1/sec^2 θ) + sin^2θ = cos^2 θ + sin^2 θ =1

3. .sin(θ - 10°) = cos64 °
sin(θ - 10°) = sin(90°-64°)
θ - 10° = 26° or 180°-(θ - 10°) = 26°
θ = 36° or θ = 164°(when0<=θ<=360°)

2007-09-01 14:50:02 補充：
1 + tan^2 θ = sec^2 θsecθ = 1/cosθ