anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you have to use \[\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\]
anonymous
  • anonymous
two of those numbers you are told you have to find the other two
anonymous
  • anonymous
do you know how to do it?

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anonymous
  • anonymous
So, if I plug in the numbers, the equetion would be\[\cos( \alpha+5)=\cos(\alpha)\cos(5)-\sin(15/17)\sin\]
anonymous
  • anonymous
oh no
anonymous
  • anonymous
I forgot to add the beta but I think you get what I'm saying, right?
anonymous
  • anonymous
lets go slow
anonymous
  • anonymous
it is not the sine of 15/17, the sine IS 15/17
anonymous
  • anonymous
lets see which two numbers you know, and which two you need
anonymous
  • anonymous
Oh, okay.
anonymous
  • anonymous
5 and 15/17 are the two numbers I have
s2rian
  • s2rian
because of \[\sin \alpha^2 + \cos \alpha^2 = 1\]
s2rian
  • s2rian
you can know other two by this equation
anonymous
  • 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
  • anonymous
i substituted directly in the formula the two numbers you were given
anonymous
  • anonymous
Oh okay. Thanks
anonymous
  • anonymous
what is missing? two more numbers, namely \[\cos(\alpha)\] and \[\sin(\beta)\]
anonymous
  • anonymous
And how exactly do I find those?
anonymous
  • anonymous
you are told that \[\sin(\alpha)=\frac{15}{17}\] and you need \(\cos(\alpha)\) do you know how to find it?
anonymous
  • 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
  • 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
  • anonymous
|dw:1450147102171:dw|solve for \(a\) using pythagoras , or from memory
anonymous
  • anonymous
Okay ignore what I said
anonymous
  • anonymous
yeah what you said is right
anonymous
  • anonymous
I see what you're saying
anonymous
  • anonymous
just need the missing side
anonymous
  • anonymous
Oh it is? Cool
anonymous
  • anonymous
let me know what you get
anonymous
  • anonymous
I got 8
anonymous
  • anonymous
Is that right or am I epically failing right now?
anonymous
  • anonymous
yes so\[\cos(\alpha)=\frac{8}{15}\]
anonymous
  • anonymous
|dw:1450147260927:dw|
anonymous
  • anonymous
\[\cos(\alpha+\beta)=\frac{8}{17}\times\frac{5}{13}-\frac{15}{17}\times \sin(\beta)\]
anonymous
  • anonymous
Got it. So now I need the exact value of cos(alpha+beta)
anonymous
  • anonymous
Okay you're ahead of me
anonymous
  • anonymous
ok i made a typo, should be \[\cos(\alpha)=\frac{8}{\color{red}{17}}\]
anonymous
  • anonymous
Alright. And then I cross multiply, right?
anonymous
  • anonymous
now find \[\sin(\beta)\] in a similar way
anonymous
  • anonymous
there is no such thing as "cross multiply" you need one more number
anonymous
  • anonymous
Oh yeah I just saw that. Sorry about that I'm so lost on this
anonymous
  • anonymous
it is just right triangle is all pythagoras all the way
anonymous
  • anonymous
So, since sin alpha is 15/17 I need to draw another triangle with sides that equal 15 and 17, right?
anonymous
  • anonymous
Wait
anonymous
  • anonymous
we got that one already
anonymous
  • anonymous
\[\cos(\alpha+\beta)=\frac{8}{17}\times\frac{5}{13}-\frac{15}{17}\times \sin(\beta)\]what number is missing?
anonymous
  • anonymous
So we do the one with the sides of 5 and 13
anonymous
  • anonymous
sin beta
anonymous
  • anonymous
yes
anonymous
  • anonymous
Okay I'll draw that
anonymous
  • anonymous
|dw:1450147660023:dw|
anonymous
  • anonymous
Would it look like that?
anonymous
  • anonymous
yes
anonymous
  • anonymous
And the missing side is 12
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[\cos(\alpha+\beta)=\frac{8}{17}\times\frac{5}{13}-\frac{15}{17}\times \frac{12}{13}\]
anonymous
  • anonymous
So the fraction would be 12/13
anonymous
  • anonymous
yes
anonymous
  • anonymous
Oh you're one step ahead of me
anonymous
  • anonymous
yes
anonymous
  • anonymous
And then do I simplify the equation or....?
anonymous
  • anonymous
there is no such mathematical operation as "simplify" compute that number is what you need to do
anonymous
  • anonymous
And how exactly do I do that?
anonymous
  • anonymous
i will leave the arithmetic to you the denominators are evidently the same, so you can easily subtract
anonymous
  • anonymous
Okay thank you
anonymous
  • anonymous
It'd be -140/221, correct?

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