# C3 Trig Proof Help!!!Watch

Thread starter 4 years ago
#1
Hi, I'm struggling with 2 questions asking to prove trig identities.

1) Prove sec2x (x) cosec2x= 2 + cot2x + tan2x

I've started on the right hand side and wrote 2sin2x + 2cos2x + (cos2x)(/sin2x) + (sin2x)/(cos2x), and then split up these fractions to get 3/sin^2x + 3/cos^x = 3cosec^2x + 3sec^2x = cosec^2x +sec^2x (however, the proof says x not +), what have I done wrong? How does the times sign appear? Could someone please help

2) Prove sec4x - tan4x = 2sec2x - 1

As for this question, I have no idea where to start since I get nowhere near the correct proof, could someone give me a hand?

Thank you!!!
0
4 years ago
#2
(Original post by CleverGirl383)
Hi, I'm struggling with 2 questions asking to prove trig identities.

1) Prove sec2x x cosec2x= 2 + cot2x + tan2

I've started on the right hand side and wrote 2sin2x + 2cos2x + (cos2x)(/sin2x) + (sin2x)/(cos2x), and then split up these fractions to get 3/sin^2x + 3/cos^x = 3cosec^2x + 3sec^2x = cosec^2x +sec^2x (however, the proof says x not +), what have I done wrong? How does the times sign appear? Could someone please help

2) Prove sec4x - tan4x = 2sec2x - 1

As for this question, I have no idea where to start since I get nowhere near the correct proof, could someone give me a hand?

Thank you!!!
1st
try starting from LHS

2nd
Can you see the difference of squares in the LHS?
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4 years ago
#3
(Original post by TeeEm)
1st
try starting from LHS

2nd
Can you see the difference of squares in the LHS?
Mm im also stuck with the first one. Ive done the second one. How should I start the first one? I tried (1+tan^2x)(1+cot^2x) but don't seem to be getting anywhere. Any hints on where to start?

Edit: never mind I didn't see the 2 at the beginning oops
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4 years ago
#4
(Original post by Super199)
Mm im also stuck with the first one. Ive done the second one. How should I start the first one? I tried (1+tan^2x)(1+cot^2x) but don't seem to be getting anywhere. Any hints on where to start?
multiply out and note cotx = 1/tanx
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Thread starter 4 years ago
#5
For Q1, I've tried using LHS and got up to 1/cos2x x 1/sin2x = 1/cos2xsin2x = sin2x + cos2x /cos2xsin2x, but I'm getting tan2x + cot2x, and not the +2... what have I done?

As for Q2, (sec2x + tan2x)(sec2x - tan2x) is the difference of squares here, but how does that help me?
0
4 years ago
#6
(Original post by CleverGirl383)
For Q1, I've tried using LHS and got up to 1/cos2x x 1/sin2x = 1/cos2xsin2x = sin2x + cos2x /cos2xsin2x, but I'm getting tan2x + cot2x, and not the +2... what have I done?

As for Q2, (sec2x + tan2x)(sec2x - tan2x) is the difference of squares here, but how does that help me?
look carefully at the second bracket
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Thread starter 4 years ago
#7
(Original post by TeeEm)
look carefully at the second bracket
I've wrote the second bracket as 1/cos^2x - sin^2x / cos^2x = 1-sin^2x / cos^2x = 1/cos^2x - sin^2x / cos^2x, getting an answer of sec^2x - 1. But I'm still somehow missing a 2 at the front of the sec term?
0
4 years ago
#8
(Original post by CleverGirl383)
I've wrote the second bracket as 1/cos^2x - sin^2x / cos^2x = 1-sin^2x / cos^2x = 1/cos^2x - sin^2x / cos^2x, getting an answer of sec^2x - 1. But I'm still somehow missing a 2 at the front of the sec term?
isnt the second bracket just 1?
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Thread starter 4 years ago
#9
(Original post by TeeEm)
isnt the second bracket just 1?
Yeah it is 1... So that leaves just (sec2x + tan2x)? do I have to set this to equal to 0? I tried to write it as 1/cos^2x + sin^2x / cos^2x and got 1+sin^2x / cos^2x, final answer was sec^2x + 1 (not the correct proof!), these are so confusing..
0
4 years ago
#10
(Original post by CleverGirl383)
Yeah it is 1... So that leaves just (sec2x + tan2x)? do I have to set this to equal to 0? I tried to write it as 1/cos^2x + sin^2x / cos^2x and got 1+sin^2x / cos^2x, final answer was sec^2x + 1 (not the correct proof!), these are so confusing..
You do not set to 0 !!!!!

Use the obvious identity. You are line away from your answer!
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Thread starter 4 years ago
#11
(Original post by TeeEm)
You do not set to 0 !!!!!

Use the obvious identity. You are line away from your answer!
If I use an identity I can write 1 as sec2x - tan2x, so I end up with 2sec2x -tan2x (don't know where the -1 comes from?!)
0
4 years ago
#12
(Original post by CleverGirl383)
If I use an identity I can write 1 as sec2x - tan2x, so I end up with 2sec2x -tan2x (don't know where the -1 comes from?!)
sec2x+tan2x = sec2x +(sec2x - 1)
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Thread starter 4 years ago
#13
(Original post by TeeEm)
sec2x+tan2x = sec2x +(sec2x - 1)
Is that an identity you're supposed to know? Never seen that before.
0
4 years ago
#14
(Original post by CleverGirl383)
Is that an identity you're supposed to know? Never seen that before.
1+tan2x=sec2x

I believe the above identity is common knowledge but I could be wrong
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Thread starter 4 years ago
#15
(Original post by TeeEm)
1+tan2x=sec2x

I believe the above identity is common knowledge but I could be wrong
Ahh so that's how you derived it! Thanks very much, got the answer now for Q2! What about Question 1?
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4 years ago
#16
(Original post by CleverGirl383)
Ahh so that's how you derived it! Thanks very much, got the answer now for Q2! What about Question 1?
look at posts 2 and 4
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Thread starter 4 years ago
#17
(Original post by TeeEm)
look at posts 2 and 4
This is my method for Q1: Start LHS: 1/cos2x (x) 1/sin2x = sec2x + cosec2x --> (tan2x + 1) + (1 + cot2x) = 2 + cot2x + tan2x.

Is this correct?
0
4 years ago
#18
(Original post by CleverGirl383)
This is my method for Q1: Start LHS: 1/cos2x (x) 1/sin2x = sec2x + cosec2x --> (tan2x + 1) + (1 + cot2x) = 2 + cot2x + tan2x.

Is this correct?
this is not correct

after the red you show that multiplication is addition
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Thread starter 4 years ago
#19
(Original post by TeeEm)
this is not correct

after the red you show that multiplication is addition
Yes but whether you add or multiply, you get the exact same value, I did this using my calculator by substituting a number for x. Did you have another method?
0
4 years ago
#20
(Original post by CleverGirl383)
Yes but whether you add or multiply, you get the exact same value, I did this using my calculator by substituting a number for x. Did you have another method?
sin0 + sin90 =sin(0+90) works

but sinA + sinB does not equal to the sin(A+B)

look at post 4

read what the other chap mentioned, then read my response.
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