Trig Identities??

NL32

12

Can someone help explain this to me at national 5 level? I literally have no idea how to do it :confused:

Reply 1

L'Evil Fish

18

What level is national 5 level?

What sort of questions do they ask

Reply 2

Pennyarcade

14

Do you mean in terms of sin, cos and tan?

Reply 3

NL32

OP

12

Original post by Pennyarcade

Do you mean in terms of sin, cos and tan?

Original post by L'Evil Fish

What level is national 5 level?

What sort of questions do they ask

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

Reply 4

Pennyarcade

14

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

Reply 5

langlitz

17

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

Um no... 1-sinx =/= cosx

Reply 6

Pennyarcade

14

Original post by langlitz

Um no... 1-sinx =/= cosx

Woops, was trying to type that too quickly.. but yeah, 1-(sinx)^2 =(cosx)^2, so the answer is still valid.

Reply 7

VividBandicoot

18

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

Reply 8

Ecasx

3

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.

Reply 9

VividBandicoot

18

That's pretty cool, never noticed it like that before

Reply 10

Ecasx

3

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.

Reply 11

NL32

OP

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

Lucky you Ha, if I didn't answer this type of question in my prelim last week correctly I wouldn't have got an A

Trig Identities??

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