# Simplifying Trig Functions

• Apr 27th 2007, 02:24 PM
cuteisa89
Simplifying Trig Functions
Can someone help me with this problems?

Simplify

1. (cos^2 (pi/2 - x))/cos x. I have no idea how to solve this one
2. tan^2 x /sec^2 x . Could this one be -1/sec^2 x ?
3. 1/(tan^2 x + 1)
4. tan x - (sec^2 x )/ tan x. I have -1/tan x

Thanks!
Isabel :)
• Apr 27th 2007, 02:58 PM
cuteisa89
Another little question
I have another problem I'm stuck in.

It says to use the given values to evaluate the other four trigonometric functions.

cot x = -3 and sin x = sqrt(10)/10

I really don't know where to start :confused: ...
• Apr 27th 2007, 04:05 PM
Jhevon
Quote:

Originally Posted by cuteisa89
Can someone help me with this problems?

Simplify

1. (cos^2 (pi/2 - x))/cos x. I have no idea how to solve this one

what you have to remember for this one is that the sine of an angle is equal to the cosine of it's complement, and vice versa. two angles are compliments (or "complimentory angles") if their sum is 90 degrees. example, 60 and 30 are compliments, since 60 + 30 = 90. in radians, pi/4 and pi/4 are compliments, since pi/4 + pi/4 = pi/2 = 90 degrees.

so cos60 = sin30
cos45 = sin45
cos55 = sin35
etc

now we can think of the angles in terms of their difference from 90
example, cos60 = cos(90 - 30) = sin30
in otherwords, whatever i have to subtract from 90 to get the first angle, the sine of that angle is equal to the cosine of the first angle. that being said, what do you think cos^2(pi/2 - x) is in sine? you guessed it, sin^2x, since (pi/2 - x) + x = pi/2 = 90 degrees

so, (cos^2 (pi/2 - x))/cos x = sin^2(x)/cosx
......................................= sin(x)/cos(x) * sin(x)
......................................= sin(x)tan(x)

Quote:

2. tan^2 x /sec^2 x . Could this one be -1/sec^2 x ?
remember, sec^2(x) = 1/cos^2(x)

so, tan^2(x)/sec^2(x) = tan^2(x)/(1/cos^2(x))
................................= tan^2(x) * cos^2(x)
................................= sin^2(x)/cos^2(x) * cos^2(x)
................................= sin^2(x)

Quote:

3. 1/(tan^2 x + 1)
tan^2(x) + 1 = sec^2(x)

so, 1/(tan^2(x) + 1) = 1/sec^2(x)
.............................= 1/(1/cos^2(x))
.............................= cos^2(x)

Quote:

4. tan x - (sec^2 x )/ tan x. I have -1/tan x
i suppose the two functions are separate since you only have the sec^2(x) in brackets over the tanx

tan(x) - sec^2(x)/tan(x) = tan(x) - (1/cos^2(x))/(sinx/cosx)
..................................= tan(x) - (1/cos^2(x)) * (cosx/sinx)
..................................= tan(x) - 1/(sin(x)cos(x))
..................................= tan(x) - sec(x)csc(x)
Not sure how simple you want this one
• Apr 27th 2007, 04:23 PM
Jhevon
Quote:

Originally Posted by cuteisa89
I have another problem I'm stuck in.

It says to use the given values to evaluate the other four trigonometric functions.

cot x = -3 and sin x = sqrt(10)/10

I really don't know where to start :confused: ...

for questions like these, the easiest way i think to do them are to draw the triangles using the trig ratios.

that is, sine = opp/hyp, cosine = adj/hyp, tangent = opp/adj etc. using those rules we can draw a triangle for each as I did below. how i filled in the sides isn't immediately obvious, ask if you don't get something.

i used the appropriate negative value on the triangle, so our signs should fall in order, however you want to check the signs when you're done. here the cotangent and hence the tangent is negative, and the sine is positive, so we must be in the second quadrant, so cosine and secant will be negative also. cosecant will be positive.

let's check that our triangle is correct:
from the triangle, sin(x) = opp/hyp = sqrt(10)/10
and cot(x) = 1/tan(x) = adj/opp = -sqrt(90)/sqrt(10) = -sqrt(9)*sqrt(10)/sqrt(10) = -3sqrt(10)/sqrt(10) = -3
so we're good

so directly from our triangle we have:

cos(x) = adj/hyp = -sqrt(90)/10 = -sqrt(9)*sqrt(10)/10 = -3sqrt(10)/10
tan(x) = 1/cot(x) = -1/3
sec(x) = 1/cos(x) = -10/3sqrt(10)
csc(x) = 1/sin(x) = 10/sqrt(10)