WILL FAN AND MEDAL
Find the exact value of cos(α+β) under the given conditions. sinα=15/17, 0<α<π/2; cosβ=5/13, 0<β<π/2

- anonymous

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- anonymous

you have to use \[\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\]

- anonymous

two of those numbers you are told
you have to find the other two

- anonymous

do you know how to do it?

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## More answers

- anonymous

So, if I plug in the numbers, the equetion would be\[\cos( \alpha+5)=\cos(\alpha)\cos(5)-\sin(15/17)\sin\]

- anonymous

oh no

- anonymous

I forgot to add the beta but I think you get what I'm saying, right?

- anonymous

lets go slow

- anonymous

it is not the sine of 15/17, the sine IS 15/17

- anonymous

lets see which two numbers you know, and which two you need

- anonymous

Oh, okay.

- anonymous

5 and 15/17 are the two numbers I have

- s2rian

because of \[\sin \alpha^2 + \cos \alpha^2 = 1\]

- s2rian

you can know other two by this equation

- anonymous

\[\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\]\[\cos(\alpha+\beta)=\cos(\alpha)\times\frac{5}{13}-\frac{15}{17}\times \sin(\beta)\]

- anonymous

i substituted directly in the formula the two numbers you were given

- anonymous

Oh okay. Thanks

- anonymous

what is missing? two more numbers, namely \[\cos(\alpha)\] and \[\sin(\beta)\]

- anonymous

And how exactly do I find those?

- anonymous

you are told that \[\sin(\alpha)=\frac{15}{17}\] and you need \(\cos(\alpha)\)
do you know how to find it?

- anonymous

there are a couple ways
method one
use \[\cos^2(\alpha)+\sin^2(\alpha)=1\]
i find it simpler to draw a triangle

- anonymous

Well, sin is Opposite over Hyp and Cosine is Adj. over hypotenuse so I know that the bottom number of the fraction would be 17

- anonymous

|dw:1450147102171:dw|solve for \(a\) using pythagoras , or from memory

- anonymous

Okay ignore what I said

- anonymous

yeah what you said is right

- anonymous

I see what you're saying

- anonymous

just need the missing side

- anonymous

Oh it is? Cool

- anonymous

let me know what you get

- anonymous

I got 8

- anonymous

Is that right or am I epically failing right now?

- anonymous

yes so\[\cos(\alpha)=\frac{8}{15}\]

- anonymous

|dw:1450147260927:dw|

- anonymous

\[\cos(\alpha+\beta)=\frac{8}{17}\times\frac{5}{13}-\frac{15}{17}\times \sin(\beta)\]

- anonymous

Got it. So now I need the exact value of cos(alpha+beta)

- anonymous

Okay you're ahead of me

- anonymous

ok i made a typo, should be \[\cos(\alpha)=\frac{8}{\color{red}{17}}\]

- anonymous

Alright. And then I cross multiply, right?

- anonymous

now find \[\sin(\beta)\] in a similar way

- anonymous

there is no such thing as "cross multiply"
you need one more number

- anonymous

Oh yeah I just saw that. Sorry about that I'm so lost on this

- anonymous

it is just right triangle is all
pythagoras all the way

- anonymous

So, since sin alpha is 15/17 I need to draw another triangle with sides that equal 15 and 17, right?

- anonymous

Wait

- anonymous

we got that one already

- anonymous

\[\cos(\alpha+\beta)=\frac{8}{17}\times\frac{5}{13}-\frac{15}{17}\times \sin(\beta)\]what number is missing?

- anonymous

So we do the one with the sides of 5 and 13

- anonymous

sin beta

- anonymous

yes

- anonymous

Okay I'll draw that

- anonymous

|dw:1450147660023:dw|

- anonymous

Would it look like that?

- anonymous

yes

- anonymous

And the missing side is 12

- anonymous

yes

- anonymous

\[\cos(\alpha+\beta)=\frac{8}{17}\times\frac{5}{13}-\frac{15}{17}\times \frac{12}{13}\]

- anonymous

So the fraction would be 12/13

- anonymous

yes

- anonymous

Oh you're one step ahead of me

- anonymous

yes

- anonymous

And then do I simplify the equation or....?

- anonymous

there is no such mathematical operation as "simplify" compute that number is what you need to do

- anonymous

And how exactly do I do that?

- anonymous

i will leave the arithmetic to you
the denominators are evidently the same, so you can easily subtract

- anonymous

Okay thank you

- anonymous

It'd be -140/221, correct?

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