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Trigonometry, EXT1 T2 2021 HSC 13d

  1. The numbers  `A`,  `B`  and  `C`  are related by the equations  `A = B - d`  and  `C = B + d`,  where  `d`  is a constant.
  2. Show that  `(sin A + sin C)/(cos A + cos C) = tan B`.  (2 marks)

  3. Hence, or otherwise, solve  `(sin\ (5theta)/7 + sin\ (6theta)/7)/(cos\ (5theta)/7 + cos\ (6theta)/7) = sqrt3`  for  `0 <= theta <= 2pi`.  (2 marks)

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  1. `text(See Worked Solution)`
  2. `(14pi)/33, (56pi)/33`
Show Worked Solution
i.    `(sin A + sin C)/(cos A + cos C)` `= (sin (B – d) + sin (B + d))/(cos (B – d) + cos (B + d))`
    `= (2sin B cos d)/(2cos B cosd)`
    `= (sin B)/(cos B)`
    `= tan B`
    `=\ text(RHS)`

 

ii.   `text(Let)\ \ A = (5theta)/7,\ \ C = (6theta)/7`

♦ Mean mark 50%.
`B` `= (A + C)/2`
  `= 1/2((5theta)/7 + (6theta)/7)`
  `= (11theta)/14`

 

`tan\ (11theta)/14` `= sqrt3`
`(11theta)/14` `= pi/3, (4pi)/3`
`:.theta` `= (14pi)/33, (56pi)/33`

Trigonometry, EXT1 T3 2020 HSC 14b

  1. Show that  `sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`.  (2 marks)

  2. By letting  `x = 4sin theta`  in the cubic equation  `x^3-12x + 8 = 0`.

     

    Show that  `sin (3theta) = 1/2`. (2 marks)

  3. Prove that  `sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`.  (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(Prove:)\  \ sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`

`text(LHS)` `= sin^3 theta-3/4 sin theta + 1/4 (sin 2thetacostheta + cos2thetasintheta)`
  `= sin^3 theta-3/4 sintheta + 1/4(2sinthetacos^2theta + sintheta(1 – 2sin^2theta))`
  `= sin^3theta-3/4 sintheta + 1/4(2sintheta(1-sin^2theta) + sintheta – 2sin^3theta)`
  `= sin^3theta-3/4 sintheta + 1/4(2sintheta-2sin^3theta + sintheta-2sin^3theta)`
  `= sin^3theta-3/4sintheta + 3/4sintheta-sin^3theta`
  `= 0`

 

ii.   `text(Show)\ \ sin(3theta) = 1/2`

`text{Using part (i):}`

`(sin(3theta))/4` `= 3/4 sintheta-sin^3 theta`
`sin(3theta)` `= 3sintheta-4sin^3theta\ …\ (1)`

 
`x^3-12x + 8 = 0`

`text(Let)\ \ x = 4 sin theta`

`(4sintheta)^3-12(4sintheta) + 8` `= 0`
`64sin^3theta-48sintheta` `= 0`
`−16underbrace{(3sintheta-4sin^2theta)}_text{see (1) above}` `= −8`
`−16 sin(3theta)` `= −8`
`sin(3theta)` `= 1/2`
♦♦♦ Mean mark (iii) 21%.

 

iii.   `text(Prove:)\ \ sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`

`text(Solutions to)\ \ x^3-12x + 8 = 0\ \ text(are)`

`x = 4sintheta\ \ text(where)\ \ sin(3theta) = 1/2`

`text(When)\ \ sin3theta = 1/2,`

`3theta` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, (25pi)/6, (29pi)/6, …`
`theta` `= pi/18, (5pi)/18, (13pi)/18, (17pi)/18, (25pi)/18, (29pi)/18, …`

 
`:.\ text(Solutions)`

`x = 4sin\ pi/18 \ \ \ (= 4sin\ (17pi)/18)`

`x = 4sin\ (5pi)/18 \ \ \ (= 4sin\ (13pi)/18)`

`x = 4sin\ (25pi)/18 \ \ \ (= 4sin\ (29pi)/18)`
 

`text(If roots of)\ \ x^3-12x + 8 = 0\ \ text(are)\ \ α, β, γ:`

`α + β + γ = −b/a = 0`

`αβ + βγ + αγ = c/a = −12`

`(4sin\ pi/18)^2 + (4sin\ (5pi)/18)^2 + (4sin\ (25pi)/18)^2` `= (α + β + γ)^2 – 2(αβ + βγ + αγ)`
`16(sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18)` `= 0-2(−12)`
`:. sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18` `= 24/16=3/2`

Trigonometry, EXT1 T3 EQ-Bank 1

  1. Show that `sinx + sin3x = 2sin2xcosx`.  (2 marks)

  2. Hence or otherwise, find all values of `x` that satisfy
     
    `qquad sinx + sin2x + sin3x = 0,\ \ \ x in [0,2pi]`.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `x = 0, pi/2, pi, (3pi)/2, 2pi, (2pi)/3, (4pi)/3`
Show Worked Solution
i.   `sinx + sin3x` `= sinx + sin2xcosx + cos2xsinx`
  `= sinx + 2sinxcos^2x + (cos^2x – sin^2x)sinx`
  `= sinx + 2sinxcos^2x + cos^2xsinx – sin^3x`
  `= sinx + 3sinx(1 – sin^2x) – sin^3x`
  `= sinx + 3sinx – 3sin^3x- sin^3x`
  `= 4sinx(1 – sin^2x)`
  `= 4sinxcos^2x`
  `= 2sin2xcosx`
  `=\ text(RHS)`

 

ii.    `sinx + sin2x + sin3x` `= 0`
  `sin2x + 2sin2xcosx` `= 0`
  `sin2x(1 + 2cosx)` `= 0`

 

`text(If)\ sin2x` `= 0:`
`2x` `= 0, pi, 2pi, 3pi, 4pi`
`:.x` `= 0, pi/2, pi, (3pi)/2, 2pi`
`text(If)\ \ cosx` `= -1/2:`
`:.x` `= (2pi)/3, (4pi)/3`

Trigonometry, EXT1 T3 EQ-Bank 3

Find all values of  `theta`  that satisfy the equation  `sqrt3 cos theta = sin(2theta)`.  (3 marks)

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`theta = 90° + m xx 180°, \ 60° + m xx 360°\ or\ 120° + m xx 360°`

Show Worked Solution
`sqrt3 cos theta` `= sin(2 theta)`
`2 cos theta sin theta – sqrt3 cos theta` `= 0`
`cos theta(sin theta – sqrt3/2)` `= 0`

COMMENT: Expressing “all values” is specifically mentioned in Topic Guidance as an application of arithmetic sequences.

`cos theta = 0 \or \  sin theta = sqrt3/2`

`theta = 90°, 270°\ or\ theta = 60°, 120° \ \ \ (0°<=theta<=360°)`

`:. theta = 90° + m xx 180°, \ 60° + m xx 360°\ or\ 120° + m xx 360°`

`text(where)\ \ m\ \ text(is an arbitrary integer).`

Trigonometry, EXT1 T3 SM-Bank 2

Show that

`cos3x = 4cos^3 x - 3cosx`.  (3 marks)

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`text(See Worked Solutions)`

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COMMENT: High achieving students know the 3 variants of  `cos 2x`  back to front. Here,  `cos2x` `=cos^2x-sin^2x`  breaks the back of this problem.

`text(LHS)` `= cos(2x + x)`
  `= cos2xcosx – sin2xsinx`
  `= (cos^2x – sin^2x)cosx – 2sinxcosxsinx`
  `= cos^3x – sin^2xcosx – 2sin^2xcosx`
  `= cos^3x – (1 – cos^2x)cosx – 2(1 – cos^2x)cosx`
  `= cos^3x – cosx + cos^3x – 2cosx + 2cos^3x`
  `= 4cos^3x – 3cosx`

Trigonometry, EXT1 T3 SM-Bank 13

Find all solutions of  `tan(2theta) = -tan theta`  for  `0<= theta<=2pi`.  (3 marks)

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`theta =0, \ pi/3, \ (2pi)/3, \ pi, \ (4pi)/3, \ (5pi)/3, \ 2pi`

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`(2 tan theta)/(1 – tan^2 theta) = -tan theta`

 
`text(Let)\ \ tan theta = k,`

`(2k)/(1 – k^2)` `= -k`
`2k` `= -k(1 – k^2)`
`2k + k(1 – k^2)` `= 0`
`3k-k^3` `=0`
`k(3 – k^2)` `= 0`

 
`k(3 – k^2) = 0 \ \ =>\ \  k = 0, \ k = +- sqrt3`
 

`text(If)\ \  tan theta = 0 \ => \ theta=0, \ pi, \ 2pi`

`text(If)\ \ tan theta = +- sqrt 3 => \ theta = pi/3, \ (2pi)/3, \ (4pi)/3, \ (5pi)/3`

`:. theta =0, \ pi/3, \ (2pi)/3, \ pi, \ (4pi)/3, \ (5pi)/3, \ 2pi`

Trigonometry, EXT1 T3 SM-Bank 12

 Solve  `sin(2x) = sin x`,  for  `0<= x <=2 pi.`  (3 marks)

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`x = 0, \ pi/3, \ pi, \ (5pi)/3, \2 pi`

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`2sin x cos x = sin x`

MARKER’S COMMENT: Many students expanded `sin(2x)` and then cancelled `sin x` on both sides which lost a set of solutions!

`sinx(2cos x – 1) = 0`

`sinx = 0, cosx = 1/2`

`text(If)\ \ sinx=0 \ => \ x=0, \ pi, \ 2pi`

`text(If)\ \ cos x=1/2 \ => \ x = pi/3, \ (5pi)/3`

`:. x = 0, pi/3, pi, (5pi)/3, 2pi`

Trigonometry, EXT1 T3 SM-Bank 10

Given that  `cot(2x) + 1/2 tan(x) = a cot(x)`, calculate  `a`.  (3 marks)

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`a = 1/2`

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`1/(tan(2x)) + (tan(x))/2 = a/(tan(x)),\ \ tan(x) != 0`

`(1 – tan^2(x))/(2 tan(x)) + (tan(x))/2 = a/(tan(x))`

 
`text(If)\ \ tan(x) != 0 :`

`(1 – tan^2(x)+tan^2(x))/(2tan(x))` `=a/(tan(x))`
`1` `=2a`
`:. a` `=1/2`

Trigonometry, EXT1 T3 SM-Bank 8

Solve for `x`, given that

`x sin(x) sec(2x) = 0,\ \ 0<=x<=2pi`  (2 marks)

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`x = 0, \ pi\ \ text(or)\ \ 2pi`

Show Worked Solution
`xsin(x)sec(2x)` `= (xsin(x))/(cos(2x))=0`

 
`text(Find)\ \ x\ \ text(that satisfies:)`

`x = 0\ \ text(and)\ \ cos(2x) != 0 \ => \ x=0`

`text(or,)`

`sin(x) = 0\ \ text(and)\ \ cos(2x) != 0 \ => \ x=pi, \ 2pi`

`:. x = 0, \ pi\ \ text(or)\ \ 2pi`

Trigonometry, EXT1 T3 2009 HSC 3c

  1. Prove that  `tan^2 theta = (1 - cos 2 theta)/(1 + cos 2 theta)`  provided that  `cos 2theta != -1`.   (2 marks)

  2. Hence find the exact value of   `tan\ pi/8`.    (1 mark)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `sqrt 2\ – 1`
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i.    `text(Prove)\ \ tan^2 theta = (1 – cos 2 theta)/(1 + cos 2 theta),\ cos 2 theta != -1`
`text(Using)\ \ cos 2 theta` `= 2 cos^2 theta\ – 1`
  `= 1 – 2 sin^2 theta`

 

`text(RHS)` `= (1 – (1 – 2 sin^2 theta))/(1 + (2 cos^2 theta\ – 1))`
  `= (2 sin^2 theta)/(2 cos^2 theta)`
  `= tan^2 theta\ text(… as required.)`

 

ii.    `tan^2\ pi/8` `= (1 – cos (2 xx pi/8))/(1 + cos (2 xx pi/8))`
    `= (1 – 1/sqrt2)/(1 + 1/sqrt2) xx sqrt2/sqrt2`
    `= (sqrt2 – 1)/(sqrt2 + 1) xx (sqrt 2 – 1)/(sqrt2 – 1)`
    `= (sqrt 2 – 1)^2/(2\ – 1)`
    `= (sqrt 2 -1)^2`

 
`:.\ tan\ pi/8 = sqrt 2\ – 1\ \ \ \ \ (tan(pi/8)>0)`

`text{(}tan\ pi/8 = sqrt (3\ – 2 sqrt 2)\ \ text{is also a correct answer)}`