You are Here: Home

# mathematical induction. watch

1. im stuck on proving an identity!
Attached Images

2. i have the answer....but i dont know how to type it all out...its too long. i'll try scanning it or something....hold on k? may take a while!
3. (Original post by IntegralAnomaly)
im stuck on proving an identity!
I'm actually quite shocked, that you actually need help.
4. (Original post by Bhaal85)
I'm actually quite shocked, that you actually need help.
? I always need help,thats how i learn
5. (Original post by IntegralAnomaly)
? I always need help,thats how i learn
Is this like P5 or P6 level?
6. (Original post by Bhaal85)
Is this like P5 or P6 level?
edexcel p6.
7. (Original post by IntegralAnomaly)
edexcel p6.
Sorry, I can't help you, I'm on the OCR board and have only gone up to P3.
8. (Original post by Bhaal85)
Sorry, I can't help you, I'm on the OCR board and have only gone up to P3.
Its oke,thanks anyway.
9. Have u asked on Nrich?
10. (Original post by JamesF)
no.I don't think they reply that much on there nowadays,maybe due to exams.
11. there's supposed to be side working where there is a purple star.

* tan(A-B)= tanA-tanB/(1+tanAtanB)
therefore 1 + tanAtanB= tanA-tanB/tan(A-B)

hope this helps!
Attached Images

12. okay so you have got:

tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1.

Now consider tan(k+1)x cot(x) + 1 = (tan(k+1)x + tan(x))/tan(x).

Now tan(k+1)x+tan(x) = tan(k+2)x (1-tan(x)tan(k+1)x).

Hence (tan(k+1)x + tan(x))/tan(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x.

Now subtituting tan(k+1)x cot(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 gives:

tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1. = tan(k+1)xtan(k+2)x + tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 - k - 1 = tan(k+2)xcot(x) - (k+1) - 1.
13. (Original post by p8224)
there's supposed to be side working where there is a purple star.

* tan(A-B)= tanA-tanB/(1+tanAtanB)
therefore 1 + tanAtanB= tanA-tanB/tan(A-B)

hope this helps!
thanks.
14. (Original post by theone)
okay so you have got:

tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1.

Now consider tan(k+1)x cot(x) + 1 = (tan(k+1)x + tan(x))/tan(x).

Now tan(k+1)x+tan(x) = tan(k+2)x (1-tan(x)tan(k+1)x).

Hence (tan(k+1)x + tan(x))/tan(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x.

Now subtituting tan(k+1)x cot(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 gives:

tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1. = tan(k+1)xtan(k+2)x + tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 - k - 1 = tan(k+2)xcot(x) - (k+1) - 1.
and thanks to u 2.
15. (Original post by IntegralAnomaly)
im stuck on proving an identity!
one word.
yuck.
16. (Original post by kimoni)
one word.
yuck.
u mean beautiful.
17. (Original post by IntegralAnomaly)
u mean beautiful.
only when you know how!!
18. (Original post by theone)
okay so you have got:

tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1.

Now consider tan(k+1)x cot(x) + 1 = (tan(k+1)x + tan(x))/tan(x).

Now tan(k+1)x+tan(x) = tan(k+2)x (1-tan(x)tan(k+1)x).

Hence (tan(k+1)x + tan(x))/tan(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x.

Now subtituting tan(k+1)x cot(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 gives:

tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1. = tan(k+1)xtan(k+2)x + tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 - k - 1 = tan(k+2)xcot(x) - (k+1) - 1.
I was about to say that.
19. (Original post by Bhaal85)
I was about to say that.
yea,and i knew that but i was just testing u people

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 2, 2004
Today on TSR

### Summer Bucket List is Back!

Start yours and you could win £150!

### GCSE Physics Equations sheet!

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE