maths questions

2012-07-27 9:15 pm
a in all questions starts for theta

simplify:

a) sin^2(90-a)cos(270-a) / tan^2(270+a)

b)sin(90+a)/cos(180-a) - cos(360-a)/tan(270+a) - sin a

solve:

a) 2sin a + 3cos a = 0

b) 2sin a= - (3^1/3) tan a

c) 4^x+2^(2x+1)=48

d) 5^x-5^(x-1)=60+8(5^(x-2))

simplify with + index :

a) (x^-4/3*y^2/5)^-1/2 / (x^7/12 * y^-1/5)^4

b) √ (x^5*√ (x^3*√ x))

回答 (1)

2012-07-27 11:15 pm
✔ 最佳答案
simplify :

a)
sin²(90° - a) cos(270° - a) / tan²(270° + a)
= cos²a (-sin a) / [sin²(270° + a) / cos²(270° + a)]
= cos²a (-sin a) sin²a / cos²a
= -sin³a

b)
[sin(90° + a) / cos(180° - a)] - [cos(360° - a) / tan(270° + a)] - sin a
= [cos a/ (-cos a)] - [cos a / (-cot a)] - sin a
= -1 + [cos a / (cos a / sin a)] - sin a
= -1 + sin a - sin a
= -1


=====
solve:

a)
2sin a + 3cos a = 0
2sin a = -3cos a
sin a / cos a = -3/2
tan a = -3/2
a = (180 - 56.31)°, (360 - 56.31)°
a = 123.69°, 303.69°

b)
2sin a = -[3^(1/3)]tan a
2sin a = -[3^(1/3)](sin a / cos a)
2sin a cos a = -[3^(1/3)]sin a
2sin a cos a + [3^(1/3)]sin a = 0
sin a [2 cos a + 3^(1/3)] = 0
sin a = 0 or cos a = -3^(1/3) / 2
a = 0°, 180°, 360° or a = (180 - 43.85)°, (180 + 43.85)°
a = 0°, 136.15°, 180°, 223.85°, 360°

If the equation is : 2sin a = -[3^(1/2)]tan a
The answers will be : a = 0°, 150°, 180°, 210°, 360°

c)
4^x + 2^(2x+1) = 48
4^x + 2*(2^2x) = 48
4^x + 2*(4^x) = 48
3*(4^x) = 48
4^x = 16
4^x = 4^2
x = 2

d)
5^x - 5^(x - 1) = 60 + 8[5^(x - 2)]
5^2[5^(x - 2)] - 5[5^(x - 2)] = 60 + 8[5^(x - 2)]
25[5^(x - 2)] - 5[5^(x - 2)] - 8[5^(x - 2)] = 60
(25 - 5 - 8)[5^(x - 2)] = 60
12[5^(x - 2)] = 60
5^(x - 2) = 5
x - 2 = 1
x = 3


=====
simplify with + index :

a)
[x^(-4/3) * y^(2/5)]^(-1/2) / [x^(7/12) * y^(-1/5)]^4
= [x^(2/3) * y^(-1/5)] / [x^(7/3) * y^(-4/5)]
= x^[(2/3) - (7/3)] * y^[(-1/5) - (-4/5)]
= x^(-5/3) * y^(3/5)
= y^(3/5) / x^(5/3)

b)
√[x^5 * √(x^3 * √x)]
= √[x^5 * √(x^3 * x^(1/2))]
= √[x^5 * √(x^(7/2))]
= √[x^5 * (x^(7/2))^(1/2)]
= √[x^5 * x^(7/4)]
= √[x^(27/4)]
= [x^(27/4)]^(1/2)
= x^(27/8)
參考: micatkie


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