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Trig Identities??
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NL32
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#1
Can someone help explain this to me at national 5 level? I literally have no idea how to do it 
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L'Evil Fish
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#2
Pennyarcade
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NL32
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#4
(Original post by Pennyarcade)
Do you mean in terms of sin, cos and tan?
Do you mean in terms of sin, cos and tan?
Yeah, although in some questions you have to substitute and I have no idea what to do?
There's a question im stuck on right now
http://imgur.com/D6b6vkL
And 2) show that sin x cos^2 x + sin^3 x = sin x
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Pennyarcade
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#5
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#5
1) One of the identities you need to learn is that tanx=sinx/cosx. If you square both sides of this relationship, youll see that (tanx)^2 = (sinx)^2/(cosx)^2
In the question, we have this relationship, except the denominator is (1-(sinx)^2).
Again, you need to learn that 1-sinx = cosx. (see the first line of my Q2 answer). If you were to sqare both sides of this relationship then you would get (1-(sinx)^2) = (cosx)^2
Putting this into the original question you have (tanx)^2=(sinx)^2/(cosx)^2, in other words tanx=sinx/cosx
2) A relationship you should have learned in class is that (sinx)^2 + (cosx)^2 = 1
In this question, you just have to multiply this whole relationship by sinx to get sinx(cosx)^2 + (sinx)^3 = sinx
In the question, we have this relationship, except the denominator is (1-(sinx)^2).
Again, you need to learn that 1-sinx = cosx. (see the first line of my Q2 answer). If you were to sqare both sides of this relationship then you would get (1-(sinx)^2) = (cosx)^2
Putting this into the original question you have (tanx)^2=(sinx)^2/(cosx)^2, in other words tanx=sinx/cosx
2) A relationship you should have learned in class is that (sinx)^2 + (cosx)^2 = 1
In this question, you just have to multiply this whole relationship by sinx to get sinx(cosx)^2 + (sinx)^3 = sinx
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langlitz
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#6
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#6
(Original post by Pennyarcade)
1) One of the identities you need to learn is that tanx=sinx/cosx. If you square both sides of this relationship, youll see that (tanx)^2 = (sinx)^2/(cosx)^2
In the question, we have this relationship, except the denominator is (1-(sinx)^2).
Again, you need to learn that 1-sinx = cosx. (see the first line of my Q2 answer). If you were to sqare both sides of this relationship then you would get (1-(sinx)^2) = (cosx)^2
Putting this into the original question you have (tanx)^2=(sinx)^2/(cosx)^2, in other words tanx=sinx/cosx
2) A relationship you should have learned in class is that (sinx)^2 + (cosx)^2 = 1
In this question, you just have to multiply this whole relationship by sinx to get sinx(cosx)^2 + (sinx)^3 = sinx
1) One of the identities you need to learn is that tanx=sinx/cosx. If you square both sides of this relationship, youll see that (tanx)^2 = (sinx)^2/(cosx)^2
In the question, we have this relationship, except the denominator is (1-(sinx)^2).
Again, you need to learn that 1-sinx = cosx. (see the first line of my Q2 answer). If you were to sqare both sides of this relationship then you would get (1-(sinx)^2) = (cosx)^2
Putting this into the original question you have (tanx)^2=(sinx)^2/(cosx)^2, in other words tanx=sinx/cosx
2) A relationship you should have learned in class is that (sinx)^2 + (cosx)^2 = 1
In this question, you just have to multiply this whole relationship by sinx to get sinx(cosx)^2 + (sinx)^3 = sinx
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Pennyarcade
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#7
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#7
(Original post by langlitz)
Um no... 1-sinx =/= cosx
Um no... 1-sinx =/= cosx
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VividBandicoot
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#8
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#8
Honestly, I never bothered to learn these identifies when I did nat 5 and I got an A, they do come up in higher though so it might be worth it to make sure you understand them now.
Posted from TSR Mobile
Posted from TSR Mobile
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Ecasx
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#9
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#9
It's two identities. It isn't hard.
Quick, basic proofs might interest you.
sinx = opp/hyp by definition
cosx = adj/hyp by definition
sinx/cosx = (opp/hyp) / (adj/hyp) = (opp/hyp) x (hyp/adj)
hyp terms cancel out
so sinx/cosx = opp/adj which is the definition of tanx
Hence tanx = sinx/cosx
from Pythagoras, opp2 + adj2 = hyp2
divide both sides by hyp2
(opp2/hyp2) + (adj2/hyp2) = 1
since sinx = opp/hyp, the first term here is sin2x
since cosx = adj/hyp, the second term here is cos2x
Hence sin2x + cos2x = 1
This is the most fundamental trig identity. If you do Advanced Maths/Mechanics it will arise a lot (also in Higher). Best to get your head around it now.
Quick, basic proofs might interest you.
sinx = opp/hyp by definition
cosx = adj/hyp by definition
sinx/cosx = (opp/hyp) / (adj/hyp) = (opp/hyp) x (hyp/adj)
hyp terms cancel out
so sinx/cosx = opp/adj which is the definition of tanx
Hence tanx = sinx/cosx
from Pythagoras, opp2 + adj2 = hyp2
divide both sides by hyp2
(opp2/hyp2) + (adj2/hyp2) = 1
since sinx = opp/hyp, the first term here is sin2x
since cosx = adj/hyp, the second term here is cos2x
Hence sin2x + cos2x = 1
This is the most fundamental trig identity. If you do Advanced Maths/Mechanics it will arise a lot (also in Higher). Best to get your head around it now.
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VividBandicoot
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Ecasx
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#11
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#11
Yeah, I remember being told this and it seemed so random/pointless. They are very simple derivations. It's a shame teachers don't show them.
In reality, maths is NOT a catalogue of facts to memorise.
In reality, maths is NOT a catalogue of facts to memorise.
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NL32
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#12
(Original post by Zain-A)
Honestly, I never bothered to learn these identifies when I did nat 5 and I got an A, they do come up in higher though so it might be worth it to make sure you understand them now.
Posted from TSR Mobile
Honestly, I never bothered to learn these identifies when I did nat 5 and I got an A, they do come up in higher though so it might be worth it to make sure you understand them now.
Posted from TSR Mobile
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