# How do you find the domain of g(t)=ln(cos(t)) and g(t)=(sqrt(1-e^t))?

can someone explain how to find the domain of g(t)=ln(cos(t)) and g(t)=(sqrt(1-e^t))?

Thanks in advance

### 2 Answers

- Tony RBLv 78 years agoFavorite Answer
===== g(t)=ln(cos(t))

First, the domain of

cos(t)

is all real numbers

the range is [ -1, +1 ]

The domain of

ln(x)

is all real numbers > 0

as ln(x) is not defined for numbers less than or equal to 0

for g(t) = ln(cos(t))

as the cos function is embedded in the ln function, the range of the cos function becomes the domain of the ln function

the cos function range is [ -1, +1 ] but we have to shrink that down to

( 0, +1 ]

Now we have to figure out the domain from that range, as we are no longer able to use all the real numbers.

The cos function is a repeating function, and for its range to be (0,+1] that means the domain must be only numbers in half of that function. I could explain this better with a graph. The domain will end up being

-90 degrees < t <= +90 degrees

or

-pi/2 < t <= +pi/2

( -pi/2, +pi/2 ]

My answer is not complete

"notthejake" gave a more complete answer, the entire domain for this function.

===== g(t)=(sqrt(1-e^t))

the domain of

e^t

is all real numbers

The domain of

sqrt(x)

is all positive real numbers, or all numbers greater than or equal to 0

[ 0, infinity ]

as the square root function is not defined for negative numbers, or all numbers less than 0

so for (1-e^t) to be within the domain of a sqrt function,

1-e^t must be greater than or equal to 0

so

1-e^t = 0

1=e^t

ln(1) = ln(e^t) = t

0 = t

for t > 0

e^t is greater than 1

and (1-e^t) is a negative number - can't be in a sqrt function

for t < 0

e^t is less than 1 - it's a fraction

and (1-e^t) is less than 1 but greater than 0

so this domain can be in the sqrt function

g(t)=(sqrt(1-e^t))

[ - infinity, 0 ]

===== Test each one

I will leave that job to you, just select numbers at or near the ends of each domain, and some in the middle, and evaluate the functions on a calculator

===== end

- notthejakeLv 78 years ago
the domain of ln u is u > 0

domain of ln (cos t) is cos t > 0

this is true on the right half of the circle:

-pi/2 < t < pi/2

but we also must add complete revolutions, so:

-pi/2 + 2pi * n < t < pi/2 + 2pi * n (where n is an integer)

this accounts for every incidence of the right half of the unit circle where cos > 0

******

domain of sqrt(u) is u >= 0

domain of sqrt(1 - e^t) is 1 - e^t >= 0

1 >= e^t

ln 1 >= t

ln 1 = 0 => t <= 0

domain of g(t) = sqrt(1 - e^t) is t <= 0