C3 trig help please Watch

cdeu12
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#1
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I'm trying to figure out the answers to these sorts of questions:

'Given that sin A = 4/5 and cos B = 8/17, where A is obtuse and B is acute, find the value of cos (A-B)'.

I don't get the acute/obtuse bits and what this means. Because it says in the example: sin A = 4/5 and A is obtuse therefore cos A = -3/5? Must crop up when you draw it but don't get why it's negative etc.

Likewise, what happens when both are acute, e.g. if sin A = 5/13 and sin B = 8/17.... need to find cos (A + B).

I clearly never understood all of this because I messed up C3, but want to know!

Also if anyone has any links to websites with some notes explaining this in more detail I'd be really grateful.
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cdeu12
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Ok, I've just tried the second question - the one where both are acute. It's a bit messy...

Basically I drew the triangle, with the adjacent as 13 and opposite as 5. By this logic I've tried to figure out the hypotenuse and end up with 13.92838828... ... which means when I include that as part of the answer for cos A it isn't right.
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cdeu12
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anyone?
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EvenStevens
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If you draw the sin graph, and where it's at 4/5 on the y axis, you'll it's about 0.927... radians on the x.

It also hits the graph once more before it gets to \pi

Now if sinA is obtuse, it's bigger than \frac{\pi}{2} and less than \pi and where it hits the graph (which in this case would be \pi - 0.927...) is the angle of A.

So!

\sin A = \frac{4}{5}
\sin ^{-1}{\frac{4}{5}} = 0.927...\rad
\pi - 0.927... = 2.21429 \rad

Now the reason why \cos A is negative is because the cos graph goes negative between \frac{\pi}{2} and \frac{3\pi}{2}. And if A is obtuse, then by definition, cosA must be negative.


Hope that made sense :-\
Reading through it again it's not exactly clear.

Oh, and just incase, your expansions of cos(A-B) and cos(A+B) are, respectively:

cosAcosB + sinAsinB
cosAcosB - sinAsinB
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cdeu12
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thanks for that explanation! So going by that, I'm still a bit confused as to what I've done in post #2 and how to rectify it.
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EvenStevens
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What's the exact question?
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cdeu12
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Just if sin A = 5/13 and sin B = 8/17, where A and B are acute, find the values of:

a) cos (A + B) b) sin (A - B) c) tan (A+B)

But I only really need help with what went wrong with a), considering I should be able to figure out b) and c) afterwards.

For a), basically I drew the triangle, with the adjacent as 13 and opposite as 5. By this logic I've tried to figure out the hypotenuse and end up with 13.92838828... ... which means when I include that as part of the answer for cos A it isn't right.
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EvenStevens
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Well the first thing you're doing wrong is that if sin A = \frac{5}{13}, then the opposite angle is 5 and the hypotenuse is 13. Not the adjacent.

Soh
Cah
Toa

Use it, it's helpful :P
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cdeu12
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Omg that's awwwful - I feel silly. - Thank you!
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cdeu12
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What about with a question like this:

If cos A = 15/17 and sin B = 20/29, where A is reflex and B is obtuse, find the values of

c) cot (A - B)...



Firstly I'm unsure about the reflex/obtuse bit when I've only just gotten used to acute --> obtuse. And then a cot's been shoved in there, so should I just do tan (A - B) and then 1 divided by that answer?
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EvenStevens
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Okay, reflex is anything past \pi (or 180 degrees).

The expansion of \cot (A-B) is \frac{1}{tan(A-B)} which is \frac{1}{(\frac{tanA-tanB}{1+tanAtanB})}

And to get that to look like a more reasonable fraction, you can just flip the bottom fraction over and get rid of the 1.

So it's just the same as
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
{\frac{1+tanAtanB}{tanA-tanB}
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