Given y = sin^2 x + sin x cos x
(a) Express y in the form p + r sin (2x - a), where r > 0 and a is a acute angle.
(b) Hence, find the max and min value of y.
a.maths* trigo
2007-12-07 7:38 am
回答 (2)
2007-12-07 8:09 am
✔ 最佳答案
y=sin^2 x + sin x cos x=(1/2)(1-cos2x)+(1/2)sin2x=1/2+(1/2)(sin2x-cos2x)=1/2+(√2/2)sin(2x-45)
-1≦sin(2x-45)≦1
(1-√2)/2≦sin(2x-45)≦(1+√2)/2
參考: me
2007-12-07 8:15 am
(sin x)^2 = ( 1 - cos 2x ) / 2
sin x cos x = ( sin 2x ) /2
So,
y
= (1 + sin 2x - cos 2x ) /2
= 1/2 + ( root2 / 2) sin (2x - pi/4)
max value of y = 1/2 + root2 /2 = ( 1 + root2 ) /2
min value of y = ( 1 - root2 ) /2
sin x cos x = ( sin 2x ) /2
So,
y
= (1 + sin 2x - cos 2x ) /2
= 1/2 + ( root2 / 2) sin (2x - pi/4)
max value of y = 1/2 + root2 /2 = ( 1 + root2 ) /2
min value of y = ( 1 - root2 ) /2
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