# P6 proof by inductionWatch

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#1
hi
In the process of finally nearing the end of further maths syllabus! Have got the general idea of proof by induction but can't do any of the trigonometric ones - can't see how to rearrange them.

If any1 could show me the working out wld be grand, here are a few (of the many!) examples I can't do:

A-----------SUMMATION SIGN cos(2r-1)x = (sin nx cos nx)/sinx
FROM r=1 to n
B-----------SUMMATION SIGN tan rx tan (r+1)x = tan (n+1)x cotx-n-1
FROM r=1 to n
C-----------SUMMATION SIGN FROM r=1 to 2n
r^3 = n^2 (2n+1)^2

Thanks.
0
13 years ago
#2
C-SUMMATION SIGN FROM r=1 to 2n
r^3 = n^2 (2n+1)^2
Assume it is true for n=k: Sum from 1 to 2k of r^3 = k^2(2k+1)^2
For n=k+1: Sum from 1 to 2k+2 of r^3 = k^2(2k+1)^2 + (2k+1)^3 + (2k+2)^3 [Don't forget that you have to add two terms, that in 2k+1 and that in 2(k+1)] [Remember that we are trying to show it is true for n=k+1 so we need to show our statement is equal to (k+1)^2(2k+3)^2.]
=k^2(2k+1)^2 +(2k+1)^3+8(k+1)^3
=(2k+1)^2[k^2+2k+1]+8(k+1)^3
=(2k+1)^2(k+1)^2+8(k+1)^3
=[(k+1)^2][(2k+1)^2+8(k+1)]
=(k+1)^2[4k^2+12k+9]
=(k+1)^2(2k+3)^2
=(k+1)^2[2(k+1)+1]^2.
But this is the same result as that obtained using the statement we are trying to prove. Hence if it is true for n=k then it is also true for n=k+1.
For n=1: Sum to 2 from 1 of r^3 = 2^3 + 1^3 = 9.
Using statement: 1^2(3^2)=9.
So the statement is true for n=1 and hence also true for n=1+1=2, 2+1=3, 3+1=4.... and so on for all positive integral n.
0
13 years ago
#3
For 2:
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
=> tanAtanB = [(tanA-tanB)/tan(A-B)] - 1

So your summation is equal to:
[(tan(r+1)x-tan(rx))/tanx] - 1

Assume it's true for n=k:
f(k) = tan (k+1)x cotx - k - 1

Let n=k+1:
f(k+1) = tan(k+1)x cotx - k - 1 + [(tan(k+2)x-tan(k+1)x)/tanx] - 1
= tan(k+1)x cotx - k - 1 - 1 + (tan(k+2)x/tanx) - tan(k+1)x/tanx
= tan(k+1)x cotx - (k+1) - 1 + tan(k+2)x cotx - tan(k+1)x cotx
= tan(k+2)x cotx - (k+1) - 1

So it holds for n=k+1 iff it holds for n=k. Let's check n=1:
f(1) = (tan2x-tanx)/tanx - 1
= (tan2x/tanx) - 1 - 1
= tan2x cotx - 1 - 1
So it holds for n=1.

Thus, by induction, it holds for all non-zero integral n.
0
#4
Thankyou - I hadn't put those two extra terms in.
0
#5
Thanks dvs!
0
#6
thanks to Jonny W too,
0
13 years ago
#7
For A:
Assume it is true for n=k: Sum from 1 to k of cos(2r-1)x=sinkxcoskx/sinx.
Hence the sum from 1 to k+1 of cos(2r-1)x=sinkxcoskx/sinx + cos(2k+1)x
=[sinkxcoskx+cos(2k+1)xsinx]/sinx
Now, sin(A+B)+sin(A-B)=2sinAcosB so sinAcosB=0.5[sin(A+B)+sin(A-B)]
sinkxcoskx=0.5[sin2kx+sin0]=0.5sin2kx [A=B=kx]
sinxcos(2k+1)x=0.5[sin2(k+1)x+sin(-2kx)]=0.5sin2(k+1)x-0.5sin2kx[Sin is an even function] [A=x, B=(2k+1)x]
So sinkxcoskx+sinxcos(2k+1)x / sinx = 0.5sin2kx+0.5sin2(k+1)x-0.5sin2kx / sinx
=0.5sin2(k+1)x/sinx=sin(k+1)xcos(k+1)x/sinx. But this is the result obtained using the given statement so if it is true for n=k then it is also true for n=k+1.
For n=1. cos(2r-1)x=cosx.
sin1xcos1x/sinx=cosx.
So it is true for n=1 and hence also true for n=1+1=2, 2+1=3, 3+1=4.... and so on for all positive integral n.
Hence, by induction, the given statement is true.
0
13 years ago
#8
The trig ones are usually tricky. One method is to write down what you're aiming for and then work backwards. Try and meet in the middle. And make sure you truly are a beast when it comes to trig identities.
Or you could turn into one of those people who spot what to do straight away...
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