[[[[thx!20點!~~!急求救]]]三角恆等式

ANSWER
http://pstock.idv.tw/files-pstock/9030602.pdf

1. Cos x = 1/3, find the value of 1+tan2 x
Cos x= 1/3 (Calculator: 1/3 shift cos)
x = 70.52877937
1 + (tan x)2
= 1 + (2.828427125)2
= 1 + 8
= 9


2. Cos x = 4/5, find the value of cos x + tan x Sin x
Cos x = 4/5
x = 36.86989765
4/5 + (tan x * Sin x)
= 4/5 + (0.75 * 0.6)
= 4/5 + 0.45 (9/20)
= 1/1/4 or 5/4


3. Sin x = 24/25 find the value of tan x+ Cos x/ tan x + sin x
Sin x = 24/25
x = 73.73979529
tan x+ Cos x/ tan x + Sin x
= (3.428571429 + 0.28) / 3.428571429 + Sin x
= 1.081666667 + Sin x
= 2.041666667
= 49/24


4. tan x = 2/5, find the value of 2 sin x / sinx + 2 cos x
tan x = 2/5
x = 21.80140949
2 sin x / sinx + 2 cos x
= 0.742781352 / 0.371390676 + 1.856953382
= 0.3333333333...
= 1/3


5. tan x = 3/4, find the value of 3 sin x + 4 cos x / 2 sin x + 6 cos x
tan x = 3/4
x = 36.86989765
3 sin x + 4 cos x / 2 sin x + 6 cos x
= 0.6 *3 + 0.8 * 4 / 0.6 * 2 + 0.8 * 6
= 5/6


THESE QUESTIONS ARE SOLVED BY USING CALCULATOR.
x = THE UNKNOWN

1.
sin^2θ+cos^2θ=1
=>(sin^2θ+cos^2θ)/cos^2θ=1/cos^2θ
=>1+tan^2θ=1/cos^2θ=1/(1/3)^2=9//

2.
(cos^2θ+sin^2θ=1
(cos^2θ+sin^2θ)/cosθ=1/cosθ
=>cosθ+tanθ sinθ=1/cosθ=1/(4/5)=4/5//


2009-03-07 21:20:41 補充：
=>cosθ+tanθ sinθ=1/cosθ=1/(4/5)=45/4