What's the answer?/how do I do this? (algebraic manipulation)

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Jordan Inglis
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#1
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#1
https://isaacphysics.org/questions/m...1_1a&stage=all
help please
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mqb2766
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#2
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#2
As per your other post, what have you tried?
With a bit of factorization, pythagorean identity, its not that bad.
Last edited by mqb2766; 4 months ago
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Jordan Inglis
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#3
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#3
(Original post by mqb2766)
As per your other post, what have you tried?
With a bit of factorization, pythagorean identity, its not that bad.
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Jordan Inglis
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#4
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#4
(Original post by mqb2766)
As per your other post, what have you tried?
With a bit of factorization, pythagorean identity, its not that bad.
That's what I've done, I don't know how you would eliminate the 2 more terms, or how this could simplify any further. The site gives you options to use tan( ), but I don't see where sin/tan would come up
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Jordan Inglis
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#5
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#5
(Original post by mqb2766)
As per your other post, what have you tried?
With a bit of factorization, pythagorean identity, its not that bad.
On the last line, the squared sign should be applied to only the (ccos(x)-v), not the whole thing, I just wrote it out wrong
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mqb2766
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#6
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#6
(Original post by Jordan Inglis)
That's what I've done, I don't know how you would eliminate the 2 more terms, or how this could simplify any further. The site gives you options to use tan( ), but I don't see where sin/tan would come up
ok, on the top you;ve got the chance of a sin^2 and a cos^2 which should simplify?
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Jordan Inglis
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#7
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#7
(Original post by mqb2766)
ok, on the top you;ve got the chance of a sin^2 and a cos^2 which should simplify?
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Size:  167.9 KB. I did that and simplified it to c^2. I don't know enough about trig identities to know what else cancels
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mqb2766
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#8
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#8
Just use the same identity again for
-sin^2 = cos^2 - 1
which should get you there.
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Jordan Inglis
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#9
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#9
(Original post by mqb2766)
Just use the same identity again for
-sin^2 = cos^2 - 1
which should get you there.
Yeah, I got the answer, tysm!
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