Trigonometric identities question Watch

tim_72
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Given that sin x + cos x = 2/3, find cos 4x.

How would you go about answering this question?

So far I tried double angle identities.

Thank you
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Dalek1099
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First, you want to convert sinx+cosx into rsin(x+a).
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amaraub
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(Original post by Dalek1099)
First, you want to convert sinx+cosx into rsin(x+a).
how do you know when to do this?
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Anonymouspsych
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(Original post by amaraub)
how do you know when to do this?
It's pretty obvious if you get some linear combination of sines and coses. Otherwise there is no other simple method of solving the equation because you are dealing with two different trigonometric functions.
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Dalek1099
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(Original post by amaraub)
how do you know when to do this?
I'm not sure I understand your question. It's true that this isn't necessarily the first thing you might decide to do but its often one of the first things you should try when you see something like Asinx+Bcosx in an exam and is one way of doing this problem.
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the bear
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(Original post by amaraub)
how do you know when to do this?
if you have

asinx + bcosx = a number which is not zero

if the RHS is zero you can divide through by cosx to get tanx
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begbie68
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This should work. Quite nicely. There is no need to use Rcos(x-a), nor Rsin(x+a) for this question. & even if you did, I'm not too sure how that might help.
(Original post by tim_72)
Given that sin x + cos x = 2/3, find cos 4x.

How would you go about answering this question?

So far I tried double angle identities.

Thank you
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ghostwalker
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:holmes: The double angle approach works quite nicely, particularly if you square the original equation you've been given. I'm about to go out, but will look again when I'm back.
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the bear
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for this question you can do what Ghostwalker suggested. this leads directly to the value of Sin2x.

this in turn can be used to get Cos4x
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Anonymouspsych
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As ghostwalker said, it is a very nice approach to square the original equation as then (by using suitable trig identities) you can get an equation in just sin(2x). Once you have found the value of sin(2x), you can then easily find the value of cos(4x) by noting that cos(4x) = 1 - 2sin^2(2x)
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begbie68
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Nice one, Bear
(Original post by the bear)
for this question you can do what Ghostwalker suggested. this leads directly to the value of Sin2x.

this in turn can be used to get Cos4x
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Dalek1099
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(Original post by begbie68)
This should work. Quite nicely. There is no need to use Rcos(x-a), nor Rsin(x+a) for this question. & even if you did, I'm not too sure how that might help.
It works because if you calculate a you get pi/4 and then 4a is pi.
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begbie68
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Yeah. And then how do you get from Rcos(x+pi/4) = 2/3 to finding value for cos4x? oh. I get it. you're going to solve for x and then quadruple it??
Ok. so what value for cos4x do you get this way?
(Original post by Dalek1099)
It works because if you calculate a you get pi/4 and then 4a is pi.
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Dalek1099
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(Original post by begbie68)
Yeah. And then how do you get from Rcos(x+pi/4) = 2/3 to finding value for cos4x? oh. I get it. you're going to solve for x and then quadruple it??
Ok. so what value for cos4x do you get this way?
You compute cos(4x)=cos(2x+2x) and use double angle formula.
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begbie68
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So what value for cos4x do you get?
(Original post by Dalek1099)
You compute cos(4x)=cos(2x+2x) and use double angle formula.
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Dalek1099
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(Original post by begbie68)
So what value for cos4x do you get?
31/81.
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begbie68
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cool. but no need to use dbl angle formula.

can you do it in the way suggested in post #8,#9 & #10 , too?
(Original post by Dalek1099)
31/81.
Last edited by begbie68; 4 weeks ago
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Dalek1099
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(Original post by begbie68)
cool. but no need to use dbl angle formula.

can you do it in the way suggested in post #8,#9 & #10 , too?
Yeh but I feel the method in posts #8,#9 & #10 needs a bit of intuition(a sort of you either see it or you don't sort of thing) where as my method uses more standard methods but is quite a bit longer though.
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begbie68
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OK. I get your point.
Doing it your way, eg finding a value for x, then finding value for cos4x: did you KNOW that it would lead to an EXACT value before you started?

Will your calculator ALWAYS give an exact value? .. or just sometimes?
If it's only 'sometimes' , can you predict the cases where it IS exact?
(Original post by Dalek1099)
Yeh but I feel the method in posts #8,#9 & #10 needs a bit of intuition(a sort of you either see it or you don't sort of thing) where as my method uses more standard methods but is quite a bit longer though.
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Dalek1099
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(Original post by begbie68)
OK. I get your point.
Doing it your way, eg finding a value for x, then finding value for cos4x: did you KNOW that it would lead to an EXACT value before you started?

Will your calculator ALWAYS give an exact value? .. or just sometimes?
If it's only 'sometimes' , can you predict the cases where it IS exact?
I didn't find a value for x I found sin(x+pi/4) then cos(x+pi/4) up to a plus/minus sign, which will disappear when squaring,using sin^2(x+pi/4)+cos^2(x+pi/4)=1. Then, I found cos(4(x+pi/4))=cos(4x+pi)=-cos(4x) from cos(x+pi/4) using addition and double angle formulas writing cos(4x)=cos(2x+2x) and using cos(2x)=2cos^2(x)-1 replacing x with x+pi/4.
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