Thread: Solving trig equations using identity and proving identites PLEASE help!

I need some help with the domains of trig equations.
I have solved all the equations i believe with the right answers, , you can recheck it if you feel comfortable.

a. cos2x+cosx=0
ANS:
2cosx^2+cosx-1
(2cosx-1)(cosx+1)
x = 1/2 and x = -1

x = pi/3 ,5pi/3 and -pi

are these the right answers in the domain x is an element [-pi,pi) - i believe the [ means equal to and the bracket means doesnt equal

b. 2cosx^2+3sinx=0
2(1-sinx^2)+3sinx=0
2-2sinx^2+3sinx
2sinx^2-3sinx-2
(2sinx-1)(sinx+2)
sinx=1/2 sinx=-2 - cancels since -2 isnt on the unit circle
x = 30 and 150 degrees -- in the domain [0,360]


c. cos2x-cosx^2=0
2cosx^2-1-cosx^2
cosx^2-1
(cosx+1)(cosx-1)
cosx=-1 cosx= 1
x = pi and 0 ---- in the domain [-pi,pi]


i just have a hard time with these domains for the trig equations, can someone please check over my answers to see if it fits the domains, or if i have to get rid of some answers or add some answers.

Thank you in advance




proving identities.

cosx-sinx/cosx = 1-tanx

can someone prove this identity for me?

1.)

a) 5pi/3 is not in the given interval. You need to find the equivalent angle that is.

b) You factored the quadratic in sin(x) incorrectly.

c) You are missing a solution.

Identity: assuming you meant to type (cos(x) - sin(x))/cos(x) = 1 - tan(x)

Use the property of fractions then use .

well thank you for pointing out my errors but can you be more specific?

i did them to the best of my ability and thank you for the proving identites

i figured out my error in b its -2 and +1
can you please help me with the domain, telling me its just wrong doesnt help me much, sorry if that sounds rude but im having some major difficulty with the - to + domains

for a) would it be just pi/3 and pi?
and for b is it 300 degrees an 210?

For a), 5pi/3 is a good angle, you just need to subtract 2pi from it to get an equivalent angle in the given interval. Or use cos(-x) = cos(x).

For b) it would be 210° and 330°. See you want (180+30)° and (360 - 30)°.

oops i made the mistake of thinkin i was a tall triangle for a second

so it would become -pi/3, pi/3 and negative pi? since i cannot use pi since pi is not included in the domain, or would negtiave pi be negated as well.

You can use -pi, but not pi since is is not part of the interval.

thank you very much, if you dont mind me asking one more question.

If <a and <b are both in quadrant 3, sina=-3/5 cosb=-5/13

find tan(a-b)

i did tan(5-13)
(tan5-tan13)/(1+(tan5*tan13)) = (5-13)/(1+(5*13)) = -8/66 = -4/33

is this correct? it seems off

I would use the fact that sine and cosine are both negative in quadrant 3 to state:





and so:

hm thats interesting never apporached it like that how come i get a different answer using the formula

tan(a-b) = tanA-tanB / 1+tanAtanB

did i substitute the numbers wrong? tan is the hyp if im correct.

I was under the impression that any and all of these forumlas that are related are supposed to give you the same answer

The tangent function is opposite over adjacent.

oh yes... stupid mistake thank you mark you have helped me tons

Hey, glad to be of assistance!