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Proof, EXT2 P2 2022 HSC 13b

The numbers `a_(n)`, for integers `n >= 1`, are defined as
 

               `{:[a_(1)=sqrt2],[a_(2)=sqrt(2+sqrt2)],[a_(3)=sqrt(2+sqrt(2+sqrt2)) \ text{, and so on.}]:}`
 

These numbers satisfy the relation  `a_(n+1)^(2)=2+a_(n)`, for  `n >= 1`.     (Do NOT prove this)

Use mathematical induction to prove that  `a_(n)=2cos\ pi/(2^(n+1))`, for all integers  `n >= 1`.  (4 marks)

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`text{Proof (See Worked Solution)}`

Show Worked Solution

`text{Prove}\ \ a_(n)=2cos(pi/(2^(n+1)))\ \ text{for}\ \ n >= 1`

`text{If}\ \ n=1:`

`a_1=2cos((pi)/(2^2))=2xx1/sqrt2=sqrt2`

`:.\ text{True for}\ n=1.`
 


Mean mark 58%.

`text{Assume true for}\ \ n=k:`

`a_(k)= =2cos(pi/(2^(k+1)))`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ a_(k+1)= 2cos(pi/(2^(k+2)))`

`text{LHS}` `=sqrt(2+a_k)`  
  `=sqrt(2+2cos(pi/(2^(k+1)))`  
  `=sqrt(2+2 cos(2 xx (pi/(2^(k+2)))))\ \ \ text{(using}\ \ cos(2theta)=2cos^2theta-1text{)}`  
  `=sqrt(2+2(2cos^2(pi/(2^(k+2)))-1)`  
  `=sqrt(2+4cos^2(pi/(2^(k+2)))-2)`  
  `=sqrt(4cos^2(pi/(2^(k+2)))`  
  `=2cos(pi/(2^(k+2)))`  
  `=\ text{RHS}`  

 

Proof, EXT2 P2 2019 HSC 14c

  1. Show that  `cot x - cot 2x = text(cosec)\ 2x`.  (2 marks)

  2. Use mathematical induction to prove that, for all  `n >= 1`,

`sum_(r = 1)^n\ text(cosec)(2^r x) = cot x - cot(2^n x)`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
Show Worked Solution

i.    `text(Show)\ \ cot x – cot 2x = text(cosec)\ 2x`

`text(LHS)` `= (cos x)/(sin x) – 1/(tan 2x)`
  `= (cos x)/(sin x) – (1 – tan^2 x)/(2 tan x)`
  `= (cos x)/(sin x) – ((1 – (sin^2 x)/(cos^2 x))/(2 (sin x)/(cos x)))`
  `= (cos x)/(sin x) – ((cos^2 x – sin^2 x)/(2 sin x cos x))`
  `= (2 cos^2 x – cos^2 x + sin^2 x)/(2 sin x cos x)`
  `= 1/(sin 2x)`
  `= text(cosec)\ 2x`
  `= ­text(RHS)`

 

ii.    `text(Prove)\ \ sum_(r = 1)^n\ text(cosec)(2^rx) = cot x – cot 2^n x\ \ text(for)\ \ n >= 1`

`text(Show true for)\ \ n = 1:`

♦ Mean mark 45%.

`text(LHS) = text(cosec)(2x)`

`text(RHS) = cot x – cot 2x = text(cosec)(2x)\ \ text{(using part (i))}`

`:.\ text(True for)\ \ n = 1`
 

`text(Assume true for)\ \ n = k:`

`text(cosec)\ 2x + text(cosec)\ 4x + … + text(cosec)\ 2^rx = cot x – cot 2^r x`

`text(Prove true for)\ \ n = k + 1:`

`text(i.e. cosec)\ 2x + … + text(cosec)\ 2^r x + text(cosec)\ 2^(r + 1) x = cot x – cot 2^(r + 1) x`

`text(LHS)` `= cot x – cot 2^r x + text(cosec)\ 2^(r + 1) x`
  `= cot x – cot 2^r x + text(cosec)\ (2.2^r x)`
  `= cot x – cot 2^r x + cot 2^r x – cot 2^(r + 1) x`
  `= cot x – cot 2^(r + 1) x`
  `= ­text(RHS)`

 
`:.\ text(True for)\ \ n=k+1`

`:.\ text(S)text(ince true for)\ \ n=1, text(by PMI, true for integral)\ \ n>=1.`

Proof, EXT2 P2 2009 HSC 8a

  1. Using the substitution  `t = tan\­ theta/2`, or otherwise, show that
     
    `qquad cot theta + 1/2 tan\­ theta/2 = 1/2 cot\­ theta/2.`  (2 marks)

  2. Use mathematical induction to prove that, for integers  `n >= 1`,
     
    `qquad sum_(r = 1)^n 1/2^(r - 1) tan­ x/2^r = 1/2^(n - 1) cot\­ x/2^n - 2 cot x.`  (3 marks) 

  3. Show that  `lim_(n -> oo) sum_(r = 1)^n 1/2^(r - 1) tan\­ x/2^r = 2/x - 2 cot x.`  (2 marks)

  4. Hence find the exact value of

  5.  

    `qquad tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + ….`  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `4/pi`
Show Worked Solution

i.   `t=tan\ theta/2, \ \ \ sin theta=(2t)/(1+t^2),\ \ \ cos theta=(1-t^2)/(1+t^2)`

`text(Prove)\ \ cot theta + 1/2 tan\­ theta/2 = 1/2 cot\­ theta/2`

`text(LHS)` `=cot theta + 1/2 tan\ theta/2`
  `=cos theta/sin theta+ 1/2 tan\ theta/2`
  `=(1-t^2)/(2t)+t/2`
  `=(1-t^2+t^2)/(2t)`
  `=1/(2t)`
  `=1/2 cot\ theta/2`
  `=\ text(RHS)`

 

ii.   `text(If)\ \ n = 1`

`text(LHS)` `=1/2^0 tan­\ x/2^1=tan\ x/2`
`text(RHS)` `=1/2^0 cot­\ x/2 – 2 cot x`
  `=cot\ x/2 – 2 cot x`
`text{Using part (i)},\ \ 1/2 tan\ theta/2` `= 1/2 cot\ theta/2 – cot theta,`
`:.tan\ theta/2` `= cot­\ theta/2 – 2 cot theta`
`text(RHS)` `=tan\ x/2`
  `=\ text(LHS)`
`:.\ text(True for)\ \ n=1`

 

`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ sum_(r = 1)^k 1/2^(r – 1) tan­\ x/2^r = 1/2^(k – 1) cot­\ x/2^k – 2 cot x.`

`text(Prove the result true for)\ \ n = k+1`

`text(i.e.)\ \sum_(r = 1)^(k + 1) 1/2^(r – 1) tan­ x/2^r = 1/2^k cot­ x/2^(k + 1) – 2 cot x`

`text(LHS)` `=sum_(r = 1)^(k) 1/2^(r – 1) tan­­ x/2^r + 1/2^k tan­ x/2^(k + 1)`
  `=1/2^(k – 1) cot­ x/2^k – 2 cot x + 1/2^k tan­ x/2^(k + 1)`
  `=1/2^(k – 1) (cot­ x/2^k + 1/2 tan­ x/2^(k +1)) – 2 cot x,\ \ \ \ text{(Let}\ \ theta=x/2^ktext{)}`
  `=1/2^(k – 1)(cot­\ theta + 1/2 tan\ theta/2) – 2 cot x`
  `=1/2^(k – 1)(1/2 cot\ theta/2) – 2 cot x,\ \ \ \ text{(from part (i))}`
  `=1/2^k cot\ theta/2 – 2 cot x`
  `=1/2^k cot­ x/2^(k + 1) – 2 cot x`
  `=\ text(RHS)`

 

`=>text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ \ n >= 1.`

 

iii.   `lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan­ x/2^r `

`=lim_(n -> oo) (1/2^(n-1) cot­ x/2^n – 2 cot x)`

`=lim_(n -> oo) (2/x * x/2^n * 1/(tan­ x/2^n) – 2 cot x)`

`=2/x xx lim_(n -> oo) ((x/2^n)/(tan­ x/2^n)) – 2 cot x`

 

  `=>text(As)\ \ n -> oo,\ \ x/2^n=theta->0, and`

  `=>lim_(theta-> 0) (theta)/(tan­ theta) =1`

`:.lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan­ x/2^r =2/x – 2 cot x`

 

iv.   `tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + …`

`=lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan­ (pi/2)/2^r`

`=(2/(pi/2)) – 2 cot­ pi/2`

`=4/pi`

Proof, EXT2 P2 2012 HSC 16b

  1. Show that  `tan^(-1)\ x + tan^(-1)\ y = tan^(-1)((x + y)/(1 − xy))`  for  `|\ x\ | < 1`  and  `|\ y\ | < 1`.  (1 mark)

  2. Use mathematical induction to prove 
     
        `sum_(j = 1)^n\ tan^(-1)(1/(2j^2)) = tan^(-1)(n/(n + 1))`  for all positive integers `n`.  (3 marks)

  3. Find  `lim_(n → ∞) sum_(j = 1)^n\ tan^(-1)(1/(2j^2))`.  (1 mark)

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  1. `text{Proof (See Worked Solutions.)}`
  2. `text{Proof (See Worked Solutions.)}`
  3. `pi/4`
Show Worked Solution

i.   `text(Let)\ \ A=tan^(-1)\ x,  and  B=tan^(-1)y`

`=> tan\ A=x,  and  tan\ B=y`

 

`tan\ (A+B)` `=(tan A + tan B)/(1-tanAtanB)`
  `=(x+y)/(1-xy)`
`:. A+B` `= tan^(-1)\ ((x+y)/(1-xy))\ \ \ text(… as required)`

 

ii.  `sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ (n/(n + 1))`

`text(If)\ \ j=1`

`text(LHS)` `=tan^(-1)\ (1/2)`
`text(RHS)` `=tan^(-1)\ (1/(1 + 1))`
  `=tan^(-1)\ (1/2)`
  `=\ text(LHS)`

 
`=>\ text(True for)\ \ j = 1.`

 

`text(Assume true)`

`sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ (n/(n + 1))`

`text(Need to prove)`

`sum_(j = 1)^(n + 1)\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ ((n + 1)/(n + 2))`

`text(LHS)` `=sum_(j = 1)^(n + 1)\ tan^(-1)\ (1/(2j^2))`
  `=tan^(-1)\ (n/(n + 1)) + tan^(-1)\ (1/(2(n + 1)^2))`
  `=tan^(-1)\ ((n/(n + 1) + 1/(2(n + 1)^2))/(1 − n/(n + 1) xx 1/(2(n + 1)^2))),\ \ \ text{(using part (i))}`
  `=tan^(-1)\ ((2n(n + 1)^2 + n + 1)/(2(n + 1)^3 − n))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2(n + 1)^3 − n))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2n^3 + 6n^2 + 6n + 2 − n))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2n^3 + 6n^2 + 5n + 2))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/((n + 2)(2n^2 + 2n + 1)))`
  `=tan^(-1)\ ((n + 1)/(n + 2))`
  `=\ text(RHS)`

 

`=> text(True for)\ \ j=n+1`

`:.text(S)text(ince true for)\ \ j = 1,\ text(by PMI, true for integral)\ \ j>=1`

 

iii. `lim_(n → ∞) sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2))`  `= lim_(n → ∞)\ tan^(-1)\ (n/(n + 1))`
    `= lim_(n → ∞)\ tan^(-1)\ (1/(1 + 1/n))`
    `= tan^(-1)\ 1`
    ` = pi/4`

Proof, EXT2* P2 2006 HSC 5d

  1. Use the fact that  `tan ( alpha - beta) = (tan alpha - tan beta)/(1 + tan alpha tan beta)`
    to show that
     
        `1 + tan n theta tan (n + 1) theta = cot theta (tan (n + 1) theta - tan n theta).`  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Show)`

`1 + tan n theta tan ( n + 1) theta = cot theta (tan (n+1) theta – tan n theta)`

`text(LHS)` `= underbrace{1 + tan n theta tan ( n + 1) theta}_text{rearrange part (i) identity}`
  `= (tan n theta – tan (n + 1) theta)/(tan (n theta – (n + 1) theta)`
  `= (tan n theta – tan (n + 1) theta)/(tan (-theta))`
  `=1/tan theta * – (tan n theta – tan (n + 1) theta)`
  `= cot theta (tan (n + 1) theta – tan n theta)`
  `=\ text(RHS  …  as required.)`

 

ii.   `text(Prove)\ \ tan theta tan 2 theta + tan 2 theta tan 3 theta + … + tan n theta tan (n + 1) theta`

`= -(n + 1) + cot theta tan (n + 1) theta`

`text(If)\ \ n = 1`

`text(LHS)` `= tan theta tan 2 theta`
  `= cot theta (tan 2 theta – tan theta) – 1\ \ \ text{(from (i))}`
  `= cot theta tan 2 theta – 1 – 1`
  `= -2 + cot theta tan 2 theta`
`text(RHS)` `= -2 + cot theta tan 2 theta`

`:.\ text(True for)\ \ n = 1`

 

`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ tan theta tan 2 theta + … + tan k theta tan (k + 1) theta`

`= -(k + 1) + cot theta tan (k + 1) theta`
 

`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ tan theta tan 2 theta + … + tan k theta tan (k + 1) theta + tan (k + 1) theta tan (k + 2) theta`

`= -(k + 2) + cot theta tan (k + 2) theta`

`text(LHS)` `= -(k + 1) + cot theta tan(k + 1) theta + tan (k + 1) theta tan (k + 2) theta`
  `= -(k + 1) + cot theta tan(k + 1) theta + cot theta[tan (k + 2) theta – tan (k + 1) theta] – 1`
  `= -(k + 2) + cot theta tan(k + 1) theta + cot theta tan (k + 2) theta – cot theta tan (k + 1) theta`
  `= -(k + 2) + cot theta tan(k + 2) theta`
  `=\ text(RHS)`

 
`:.\ text(True for)\ \ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 1,\ \ text(true for integral)\ \ n >= 1`