# Trig Identities??

Announcements
#1
Can someone help explain this to me at national 5 level? I literally have no idea how to do it 0
7 years ago
#2
What level is national 5 level?

What sort of questions do they ask
0
7 years ago
#3
Do you mean in terms of sin, cos and tan?
0
#4
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
0
7 years ago
#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
0
7 years ago
#6
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
0
7 years ago
#7
(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.
0
7 years ago
#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
0
7 years ago
#9
It's two identities. It isn't hard.

Quick, basic proofs might interest you.

sinx = opp/hyp by definition

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
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.
0
7 years ago
#10
That's pretty cool, never noticed it like that before
0
7 years ago
#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.
0
#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
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Have you done work experience at school/college?

Yes (155)
41.67%
Not yet, but I will soon (69)
18.55%
No (148)
39.78%