# Trig questions

1. Nov 8, 2005

### dnt

im trying to remember how to do some trig.

first of all, how do you find cos^-1 (5/3) without using a calculator?

2nd, i need help proving this identity:

2sin^2(x) - cos (x) - 2 sin (x) + 1 = 0

3rd, how do you evaluate sin 165 (which is sin (11/12 pi)) again without using a calculator.

thanks for the help. i know im suppose to try them first but i havent done this in a while and im really stuck.

also if someone could give me a good trig link so i can refresh on my trig, id appreciate it.

2. Nov 8, 2005

### benorin

sin 165 = sin(120+45), then use formula for sin(A+B)

recall that sin^2 x + cos^2 x = 1

and cos^-1 (5/3) does not exist since -1 <= cos(x) <= 1 for any x

3. Nov 8, 2005

### HallsofIvy

Staff Emeritus
As benorin said, cos(5/3) is impossible. cosine of any number cannot be larger than 1.
(2) is also impossible: it is NOT an identity. In particular, for x= 45 degrees,
sin 45= cos 45= $\frac{\sqrt{2}}{2}$ so
2 sin^2(45)- cos(45)- 2sin(45)+ 1=
$$\frac{2}{2}- \frac{\sqrt{2}}{2}- \sqrt{2}+ 1= 2- 3\frac{\sqrt{2}}{2}$$
not 0.
For (3), sin(165)= sin(120+ 45)= sin(2(60)+ 45). You should know the sine and cosine of both 60 and 45 and use the sum formula and double angle formula.

4. Nov 8, 2005

### dnt

ok, so it would be:

(sin 120)(cos 45) + (cos 120)(sin 45)

correct?

the full problem was sin[2 cos^-1(5/3)]

5. Nov 8, 2005

### dnt

i know how to do it if you use sin (120 + 45), assuming my work in the post right above is correct, but just for learning sake, how do you do it using sin (2(60) + 45) - the double angle formula with the sum formula?

edit: would you say [sin (2(60))* cos(45)] + [cos (2(60))* sin (45)] and then use the double angle formula to get:

(2)(sin 60)(cos 60) * (cos 45) + (cos 60)^2 - (sin 60)^2 * (sin 45)

Last edited: Nov 8, 2005
6. Nov 8, 2005

### dnt

another identity i could use some help in:

(csc X)/(1 - cos X) = (1 - sin X)/(cos^2 X)

what i tried to do was say cos^2 X = 1 - sin^2 X

then i factored it so it would cancel with the top. leaving me with:

(csc X)/(1 - cos X) = 1/(1 + sin X)

but now im stuck.

7. Nov 8, 2005

### HallsofIvy

Staff Emeritus
The problem you have is that that is NOT an identity. Again, if X= 45 degrees, then the formula becomes
$$\frac{\sqrt{2}}{1- \frac{\sqrt{2}}{2}}= \frac{1- \sqrt{2}}{\frac{1}{2}}$$
That is NOT true! Where are you getting these problems?

8. Nov 8, 2005

### dnt

a book. well i feel better for not being able to get them :)

thanks for the help.

9. Nov 9, 2005

### dnt

oops...the actual problem was

sin[2 cos^-1(3/5)]

(not 5/3)

now can someone help me with this? thanks.

10. Nov 9, 2005

### BerryBoy

I hope this helps you,
Regards,
Sam

11. Nov 10, 2005

### dnt

12. Nov 10, 2005

### moose

In this one "first of all, how do you find cos^-1 (5/3) without using a calculator? "

It's extremely simple

the ratio between the adjacent and hypotenus can never be greater than one, because the adjacent side cannot be longer than the hypotenus.

13. Nov 10, 2005

### dnt

thanks, but that was already explained above. how do you do cos^-1 (3/5)?

14. Nov 11, 2005

### BerryBoy

Is the question actually asking you to find the value of arccos(3/5)?

Or are you trying to work it out as a step in your calculations?

Regards,
Sam

15. Nov 11, 2005

### HallsofIvy

Staff Emeritus
You said before:
"oops...the actual problem was

sin[2 cos^-1(3/5)]"

BerryBoy then showed you how to get that, without actually finding the angle. In fact there is no way to get an exact answer for the cos-1(3/5).

16. Nov 11, 2005

### dnt

sorry, the full problem is sin[2 cos^-1(3/5)]

but is there any way to find cos^-1(3/5) by itself without a calculator?

if not, how do you find the answer to the above full problem? would it be:

2(4/5)(3/5) = 24/25

is that right?

17. Nov 11, 2005

### HallsofIvy

Staff Emeritus
There is no way, other than by approximation, either by Taylor's series or by successive approximations.

Yes,you can do sin(2arccos(3\5)) just as suggested. If cos x= 3/5, sinx= 4/5 sin(2x)= 2(4/5)(3/5)= 24/25.