Maths about Trigonometry

2014-08-05 11:37 pm
1. Solve √3 + 2cosx = 0, where 0°≤ x < 360°.

2. Solve 2cosθ + 1 = 0, where 0°≤ x < 360°.

3. Solve 3tanθ - √3 = 0, where 0°≤ x < 360°.

4. Solve tan²x + 2tanx - 24 = 0 for 0°≤ x < 360°.

5. Solve 1 - cosθ= sin²θ, where 0°≤ x < 360°.

6. Solve (7sinθ + cosθ)/(4sinθ + cosθ) = 2, where 0°≤ x < 360°.

7a. Rewrite 8tanθ = 3cosθ in the form xsin²θ + ysinθ + z = 0, where x, y and z are integers.

7b. Hence, solve the equation in (a) for 0°≤ θ ≤ 360°.

8. The figure below shows the graph of y = asinθ + k, find the value of a and k.

9. Find the maximum value and minimum value of each of the following trigonometric functions.
(a) 2 + 3cos(x + 45°)
(b) (-2/3) - 2cos3x

Steps and Ans Thank you very much!

回答 (2)

2014-08-06 1:56 am
✔ 最佳答案
1.
√3 + 2 cos x = 0
2 cos x=-√3
cos x=-(√3)/2
x=150 or x=210

2.
2 cosθ + 1 = 0
2 cosθ=-1
cosθ=-0.5
θ=120 or θ240

3.
3 tanθ - √3 = 0
tanθ=(√3)/3
θ=30 or 210

4.
tan²x + 2tanx - 24 = 0
tan x=4 or tan x=-6
x=75.96375653 or x=255.9637565 or x=279.4623222 or x=99.46232221

5.
1 - cosθ= sin²θ
1-sin²θ=cosθ
cos²θ-cosθ=0
cosθ=0 or cosθ=1
θ=90 or θ=270 or θ=0

6.
(7 sinθ + cosθ)/(4 sinθ + cosθ) = 2
7 sinθ+cosθ=8 sinθ+2 cosθ
-sinθ=cosθ
tanθ=-1
θ=315 or θ=135

7 a.
8 tanθ = 3 cosθ
8 sinθ=3 cos²θ
3 cos²θ-8 sinθ=0
3(1-sin²θ)-8 sinθ=0
-3 sin²θ-8 sinθ+3=0
3 sin²θ+8 sinθ-3=0

7 b.
3 sin²θ+8 sinθ-3=0
sinθ=1/3 or sinθ=-3 (Delete)
θ=19.47122063 or 160.5287794

8.
No figure

9 a.
-1≤cos(x+45°)≤1
-3≤3 cos(x+45°)≤3
-1≤2+ cos(x+45°)≤5

9 b.
-1≤cos3x≤1
-2≤-2cos3x≤2
-(8/3)≤ (-2/3) - 2cos3x≤(4/3)

2014-08-05 18:04:57 補充:
To 土扁:
5.
1- cos 180°=2
sin² 180°=0
1- cos 180° 不等於 sin² 180°
2014-08-06 8:13 am
原來的答案沒有 180°,但最後一時糊塗加上了。

既然如此,把答案刪掉算了。

2014-08-06 00:15:47 補充:
嚴格來說,fighting gravity 全部答案可以說是錯的。

x = 150 或 x = 210 的意思是:

x = 150 radians 或 x = 210 radians

2014-08-06 00:16:43 補充:
當然,x = 0 是正確的。


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