關於sin, cos, tan

👨🏻‍🏫 學術台數學

[匿名]

2008-07-14 3:51 pm

1) Simplify the following expressions.

a. cos(180° + x) + cos(180° - x)

b. cos(126°)cos(36°) + sin(126°)sin(36°)

2) For sin( ) = -0.4321, 0 < < 360°

a. How many solutions are possible?

b. In which quadrants would you find the solutions?

c. Determine the reference angle for this equation to the nearest degree.

d. Determine all the solutions to the equation to the nearest degree.

3) Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 < x < 360°. Use exact solutions whenever possible.

a. 6cos2( ) + cos( ) - 1 = 0

b. 2 - 2sin2( ) = cos( )

更新1:

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回答 (4)

Uncle Michael

2008-07-15 1:33 am

✔ 最佳答案

1)
a. cos(180 + x) + cos(180 - x)

a. = (-cosx) + (-cosx)

a. = -2cosx

b. cos(126)cos(36) + sin(126)sin(36)

b. = cos(126 - 36)

b. = cos90

b. = 0

=====
2)
a. There are 2 possible solutions.
a. (sine is -ve in the 3rd/4th quadrants.)

b. The solutions are found in the 3rd and 4th quadrants.

c. sinθ = +0.4321

c. Refernce angle θ = 25.6

d. The solutions are found in the 3rd and 4th quadrants.

d. ( ) = 180 + 25.6 ororor ( ) = 360 - 25.6

d. ( ) = 205.6 ororor ( ) = 334.4

=====
3)
a. 6cos( ) + cos( ) - 1 = 0

a. [2cos( ) + 1] [3cos( ) - 1] = 0

a. 2cos( ) + 1 = 0 ororor 3cos( ) - 1 = 0

a. cos( ) = -1/2 ororor cos( ) = 1/3

a. ( )=180-60 or ( )=180+60 or ( )=70.5 or ( )=360-70.5

a. ( ) = 120 oorr ( ) = 240 oorr ( ) = 70.5 oorr ( ) = 289.5

b. 2 - 2sin( ) = cos( )

b. 2[sin( ) + cos( )] - 2sin( ) = cos( )

b. 2sin( ) + 2cos( ) - 2sin( ) = cos( )

b. 2cos( ) - cos( ) = 0

b. cos( ) [2cos( ) - 1] = 0

b. cos( ) = 0 ororor cos( ) =

b. ( ) = 90 oorr ( ) = 270 oorr ( ) = 60o oorr ( ) = 360-60

b. ( ) = 90 oorr ( ) = 270 oorr ( ) = 60o oorr ( ) = 300
=

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2008-07-14 4:35 pm