Trigonometric Identities help?

Question

the directions say- solve for values of x between 0 degrees and 90 degrees.

1.) sec x = 2 , find tan x

i got tan x= root 3... but how do i convert this to degrees, and am i even doing it right?

2.) cot x= 1/2. find sin x

i got 1/sin x = root5 / 2

again....degrees??

help pleasee!!!

Answer 1

1.) sec x = 2
=> 1/cos x = 2
=> cos x = 1/2
=> tan x = √ 3
you don't need this.

1.) sec x = 2
=> 1/cos x = 2
=> cos x = 1/2
=> x = cos‾¹ 1/2
=> x = 60

2.) cot x= 1/2. find sin x
=> 1/tan x = 1/2
=> tan x = 2/1 = O/A
=> sin x = 2/√5 = O/H

=> x = sin x ‾¹(2/√5)
=> x = 63.43 deg

Answer 2

tan x = sqrt(3) is correct. You can just use your calculator and get arctan(sqrt(3)) = 60 degrees. You should draw a right triangle that is 30-60 -90 degrees. Then hypotenuse is 2, short leg is 1 and long leg is Sqrt(3). The angle between the hypotenuse and the short leg is 60 degrees. So you can immediatelly see that sec 60 = 2 and tan 60 = sqrt(3).

cot x = 1/2
Draw a right triangle with short leg =1 and long leg =2. Then hypotenuse = sqrt(5) . You can see that cot x = 1/2 and sin x = 2/sqrt(5) = .8944272
arcsin (.894472) = 63.43 degrees by calculator.
arcsin is same as sin^-1.

Answer 3

Problem 1:

recall: sin^2(x) + cos^2(x) = 1

divide by cos^2(x)

sin^2(x)/cos^2(x) + 1 = 1/cos^2x)

1/cos(x) = sec(x)

tan^2(x) + 1 = sec^2(x)

tan^2(x) = sec^2(x) - 1
tan^2(x) = 4 - 1

tan(x) = +/-sqrt(3)

x = inv tan(sqrt(3))

x = 60 degrees

2) cot(x) = 1/2 find sin(x)

start as before

sin^2(x) + cos^2(x) = 1
divide by sin^2(x)

1 + cot^2(x) = 1/ sin^2(x)

1+ 1/4 = 1/sin^2(x)

5/4 = 1/sin^2(x)

sin(x) = sqrt(4/5)

x = inv sin(sqrt(4/5))

x = 63.0 degrees

Answer 4

1. tanx = sqrt(3) is correct. To convert into degrees, you must think of which value of x will give tanx=sqrt(3), (this is like solving logarithms). It's 60 degrees.

2. sinx is 2/sqrt5. Like above (or you can use a calculator), x=63.4349488...