Find the exact value of sin(W+T)

[TayTayDatDude]

Homework Statement

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)

Homework Equations

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.

 

[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.

 

[TayTayDatDude]

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.

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

 

[montoyas7940]

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

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

 

[Chaos2009]

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

 

[TayTayDatDude]

Oh ok thanks

 

[TayTayDatDude]

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

 

[montoyas7940]

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

 

[Chaos2009]

I received the same answer, 56/65.

 

[TayTayDatDude]

I must be doing something wrong then.

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

 

[montoyas7940]

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

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

 

[TayTayDatDude]

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

 

[montoyas7940]

You could use another identity.

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

 

[TayTayDatDude]

Erm.. what?

 

[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.

 

[montoyas7940]

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

it means the same thing.

 

[montoyas7940]

Yeah, Chaos' method does make it easier.

 

[Chaos2009]

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

 

[montoyas7940]

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

 

[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.

 

[Nissen, Søren Rune]

I'd start by finding out what W and T are, using arcsin(3/5) and arcos(12/13).

Once you know the values of W and T, just add them together and plug them into the sinus function.

 

[Chaos2009]

If you were to do that, you would lose precision because you would have to round the values you obtained for W and T.

 

[eliotargy]

Boys boys! To find cos(w) and sin(t) just draw the triangles! 3,4,5 and 5,12,13. Then you have them. cos(w) = 4/5; sin(t) = 5/13. Basta!

 

[Chaos2009]

If you read up, that is basically what I originally said.

 

[eliotargy]

If you read up, that is basically what I originally said.

Yeah, but I said it easier.