# Homework Help: Trig help

1. Nov 20, 2008

### TayTayDatDude

1. The problem statement, all variables and given/known data

If sinW = 3/5 and cosT = 12/13, where both W and T are acute angles, then find the exact value of sin(W+T)

2. Relevant equations

3. The attempt at a solution

I tried sin(3/5)cos(12/13) + cos(3/5)sin(12/13) = sin [(3/5) + (12/13)] but I doubt it's right.

2. Nov 20, 2008

### Chaos2009

You were close in your attempt but you made a little mistake. The identity you are looking for is:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
You should try sinWcosT + cosWsinT. You are already given sinW and cosT. You should also be able to find cosW and sinT from those two values as well.

3. Nov 20, 2008

### TayTayDatDude

Well I did do that w = x, t = y

Sin(3/5)cos(12/13) + cos(3/5)sin(12/13)
w t w t

Or maybe I don't understand you, elaborate plaese

4. Nov 20, 2008

### montoyas7940

sinW = 3/5 not sin(3/5)

I think you are trying to use sin(sinW) instead of sinW.

5. Nov 20, 2008

### Chaos2009

Okay, that's what I mean. sinW is not sin(3/5), sinW = 3/5.

6. Nov 20, 2008

### TayTayDatDude

Oh ok thanks

7. Nov 20, 2008

### TayTayDatDude

So, I got 72/65 for sin (w+t), can someone clarify please?

8. Nov 20, 2008

### montoyas7940

Unless I fat fingered it:
I got 56/65
maybe double check...

9. Nov 20, 2008

### Chaos2009

I recieved the same answer, 56/65.

10. Nov 20, 2008

### TayTayDatDude

I must be doing something wrong then.

(3/5)(12/13) + (3/5)(12/13) = 72/65

11. Nov 20, 2008

### montoyas7940

No, you don't yet have cosW or SinT.

You need to find them first from what you have been given.

12. Nov 20, 2008

### TayTayDatDude

WEll i have been trying for awhile now to figure out cosw and sint. Any hints on how to get there?

13. Nov 20, 2008

### montoyas7940

You could use another identity.

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

14. Nov 20, 2008

### TayTayDatDude

Erm.. what?

15. Nov 20, 2008

### Chaos2009

Okay, here is another way that actually is really easy for this particular problem.
$$sin W = \frac{o}{h}$$
You should know this, it is the definition of the sin function. Well, they tell you sin W = 3/5 so o = 3 and h = 5. You can then draw a right triangle with a hypotenuse of 5 and one leg of 3 and find the length of the other leg, a, using the Pythagorean theorem. Once you find a,
$$cos W = \frac{a}{h}$$
you can plug it into there along with your h that you already know.

You can use a similar method on T.

16. Nov 20, 2008

### montoyas7940

Maybe you have seen it written sinx^2 + cosx^2 = 1

it means the same thing.

17. Nov 20, 2008

### montoyas7940

Yeah, Chaos' method does make it easier.

18. Nov 20, 2008

### Chaos2009

He means
$$sin^{2}x + cos^{2}x = 1$$

19. Nov 20, 2008

### montoyas7940

Thank you, I really should use latex. I am just so slow...

20. Nov 21, 2008

### Дьявол

sin(W+T)=sinWcosT+cosWsinT
Since sinW=3/5 and cosT=12/13
$$sin(W+T)=\frac{3}{5}*\frac{12}{13}+cosW*sinT$$
We don't know cosW and sinT, so we could find them by using:
cosW=√1-sin2W
cosW=√1-9/25
cosW=4/5
using the same method
sinT=√1-cos2T
sinT=√1-144/169
sinT=√25/169
sinT=5/13
So we can plug them in the equation above:
$$sin(W+T)=\frac{3}{5}*\frac{12}{13}+\frac{4}{5}*\frac{5}{13}$$
The final answer should be 56/65.
Regards.