1. ## trigonometric identities

i have been staring at my book for the past hour waiting for divine inspiration...

you know the identities where for example sin(x) = 1 / csc(x) and cos(x) = 1 / sec(x)

HOW do i apply that stuff to for example this (directions just say "simplify"):
cos(x) / sec(x) - tan(x)

i have tried everything. i can't get this to work. the answer according to the back of the book is 1 + sin(x)

please respond this is due on wednesday and there's about 80 problems, none of which i can comprehend!!

2. Originally Posted by anonymous
i have been staring at my book for the past hour waiting for divine inspiration...

you know the identities where for example sin(x) = 1 / csc(x) and cos(x) = 1 / sec(x)

HOW do i apply that stuff to for example this (directions just say "simplify"):
cos(x) / sec(x) - tan(x)

i have tried everything. i can't get this to work. the answer according to the back of the book is 1 + sin(x)

please respond this is due on wednesday and there's about 80 problems, none of which i can comprehend!!
I presume this is really:
cos(x)/[sec(x) - tan(x)]

The way I do these is to convert everything into sine and cosine functions:
cos(x)/[1/cos(x) - sin(x)/cos(x)]

To reduce this compound fraction, multiply the numerator and denominator by cos(x):
cos^2(x)/[1 - sin(x)]

Now, sin^2(x) + cos^2(x) = 1, so cos^2(x) = 1 - sin^2(x):

[1 - sin^2(x)]/[1 - sin(x)]

Note that the numerator factors:
1 - sin^2(x) = (1 + sin(x))(1 - sin(x))

[(1 + sin(x))(1 - sin(x))]/[1 - sin(x)]

So as long as sin(x) is not equal to 1 (ie. x is not equal to (pi)/2 rad), we may cancel the common 1 - sin(x):
1 + sin(x)

-Dan

3. Originally Posted by topsquark
The way I do these is to convert everything into sine and cosine functions:
cos(x)/[1/cos(x) - sin(x)/cos(x)]
you lost me here. i get the 1/cos(x) but where are you getting sin(x)/cos(x) from? i'm guessing this is some other identity than tangent is equal to?? i think what it is is my textbook only covers the basics.

here is what i have:
sin(x) = [1/csc(x)]
cos(x) = [1/sec(x)]
tan(x) = [1/cot(x)]
csc(x) = [1/sin(x)]
sec(x) = [1/cos(x)]
cot(x) = [1/tan(x)]

cos^2(x) + sin^2(x) = 1
tan^2(x) + 1 = sec^2(x)
1 + cot^2(x) = csc^2(x)

sin(x/2) = [1-cos(x) / 2]
cos(x/2) = [1+cos(x) / 2]
tan(x/2) = [1-cos(x) / 1 + cos(x)]

sin(2x) = [2sin(x)][cos(x)]
cos(2x) = [cos^2(x)] - [sin^2(x)]
= [2cos^2(x)] - 1
= 1 - [2sin^2(x)]
tan(2x) = [(2)(tan(x))] / [1-tan^2(x)]

cos(x y) = [cos(x)][cos(y)] [sin(x)][sin(y)]
sin(x y) = [sin(x)][cos(y)] [sin(y)][cos(x)]
tan(x y) = [tan(x) tan(y)] / [1 [tan(x)][tan(y)]]

can you take a look at this and tell me what i'm missing?? because i think that's why i can't do anything with my homework.

4. Originally Posted by anonymous
you lost me here. i get the 1/cos(x) but where are you getting sin(x)/cos(x) from? i'm guessing this is some other identity than tangent is equal to?? i think what it is is my textbook only covers the basics.

here is what i have:
sin(x) = [1/csc(x)]
cos(x) = [1/sec(x)]
tan(x) = [1/cot(x)]
csc(x) = [1/sin(x)]
sec(x) = [1/cos(x)]
cot(x) = [1/tan(x)]

cos^2(x) + sin^2(x) = 1
tan^2(x) + 1 = sec^2(x)
1 + cot^2(x) = csc^2(x)

sin(x/2) = [1-cos(x) / 2]
cos(x/2) = [1+cos(x) / 2]
tan(x/2) = [1-cos(x) / 1 + cos(x)]

sin(2x) = [2sin(x)][cos(x)]
cos(2x) = [cos^2(x)] - [sin^2(x)]
= [2cos^2(x)] - 1
= 1 - [2sin^2(x)]
tan(2x) = [(2)(tan(x))] / [1-tan^2(x)]

cos(x y) = [cos(x)][cos(y)] [sin(x)][sin(y)]
sin(x y) = [sin(x)][cos(y)] [sin(y)][cos(x)]
tan(x y) = [tan(x) tan(y)] / [1 [tan(x)][tan(y)]]

can you take a look at this and tell me what i'm missing?? because i think that's why i can't do anything with my homework.
Ummmm....
tan(x) = sin(x)/cos(x) is the definition of the tangent function. It isn't an identity, it's an out and out definition.

Go back to the basic right triangle. Given an angle x, the three basic trig functions of x are
sin(x) = opposite/hypotenuse