Expert: Sherry Wallin - 12/4/2008

Question
I am having problems solving a few trigonometric identities. If you could help, I would really appreciate it!

1. Prove the identity:

(1 + secx)/secx  = sin^2x/(1 - cosx)

2. Prove the identity:

sinx(cosx(x/2))-1 = cos x

3. Solve the identity and give all solutions:

(1 - cosx)/(1 + cosx) = 3

4. Solve the identity and give all solutions:

√3 tan3x + 3 = 0

**Only the first "3" is in the square root symbol

Answer
1-  sin^2(x)/(1-cos x) = (1-cos^2(x))/(1-cos x)
(to see this just multiply top and bottom by (1-cosx)
= (cos x "+"  1)(1-cosx)/(1-cosx)
= cos x   1
= [1  "+" (1/cos x)]/(1/cos x)
= (1 "+"  sec x)/secx


I think perhaps question 2 is written wrong, because it is not an identity. To see this put in a value for x and check it. Here are 2 examples:

sin(0)(cos(0/2)) - 1 = 0 -1 = -1 which is not equal to cos 0 = 1.
sin 90(cos 45) - 1 = sqrt(2)/2 - 1 which is not equal to cos 90 = 0

To be an identity it must work for all values of x.

Definition of Identity:
Two functions f and g are said to be identically equal if f(x) = g(x) for *EVERY VALUE OF X" for which both functions are defined. Above both sinx and cosx are defined for 0 degrees and 90 degrees.

#'s 3-4 are worded wrong because you aren't looking for solving an identity you are just looking for the values of x that will make the statement true, so you are just solving an equation for x.

3- (1 - cosx)/(1 "+"  cosx) = 3 so (1 - cosx) = 3 (1 "+"  cosx)
1 - cos x = 3 "+"  3 cos x and
-cos x = 2 "+"  3 cos x  and
-2 = -4 cos x
(1/2) = cos x
cos^-1(1/2) = x
x = 60 deg, 300 deg (or - 60 deg)

4- sqrt(3)tan(3x)   3 = 0
sqrt(3)tan(3x)= -3
tan(3x) = [-3/sqrt(3)]
tan^-1[tan(3x)] = tan^-1[-3/sqrt(3)]
3x = -60
x = -20 deg = 340 deg, 40 deg

I arrived at the 40 degrees in this way. I added 180/3 degrees to the -20 degrees and so now when we multiply the 40 degrees by 3 we get 120 degrees. Or just take -60 degrees (which comes from 3x = -60) and add 180 degrees which says 3x = 120 so x = 40.

Math Prof

PS Anywhere it seems an operation sign is missing put a plus there. I put "+" in the places I saw that don't come over this site to indicate addition. I hope this is clear.