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Trigonometry, 2ADV T2 EQ-Bank 5

Solve  \(\sin x-\cos x=0 \quad-\pi \leqslant x \leqslant \pi\)   (2 marks)

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\(x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)

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  \(\sin x-\cos x\) \(=0\)
  \(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\) \(=0\)
  \(\tan x-1\) \(=0\)
  \(\tan x\) \(=1\)
  \(x\) \(=\tan ^{-1}(1)\)

 
\(\therefore x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)

Trigonometry, 2ADV T2 SM-Bank 2

Find all solutions of the equation  `2 cos theta = sqrt 3 cot theta`,  for  `0<=theta<=2pi`  (3 marks)

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`theta=pi/3, pi/2, (2pi)/3, (3pi)/2`

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`2 cos theta = sqrt 3 cot theta`

`2 cos theta – sqrt 3 cot theta` `= 0`  
`2 cos theta – sqrt 3 (cos theta)/(sin theta)` `=0`  
`(2 – sqrt 3/sin theta) cos theta` `=0`  

 
`text(If)\ \ cos theta=0,`

`theta=pi/2, (3pi)/2`
  

`text(If)\ \ 2 – sqrt 3/sin theta = 0\ \ =>\ \ sin theta = sqrt 3/2`

`theta = pi/3, (2pi)/3`

Trigonometry, 2ADV T2 SM-Bank 44 MC

The domain of the function with rule  `f(x) = 1 - sec(x + pi/4)`  is

  1. `text(all real)\ x`
  2. `{((4k - 1)pi)/4}, (text{for}\ k\ text{integer}) `
  3. `{((4k + 1)pi)/4}, (text{for}\ k\ text{integer}) `
  4. `{((2k - 1)pi)/4}, (text{for}\ k\ text{integer}) `
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`C`

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`y = sec (x)=1/cos(x)\ \ text(has asymptotes when)`

`x = −pi/2, pi/2, (3pi)/2, …`

`=> y = sec (x + pi/4)\ \ text(has asymptotes at when)`

`x = (−3pi)/4, pi/4, (5pi)/4, …`
 

`:.\ text(Domain:)\ {((4k + 1)pi)/4}, (text{for}\ k\ text{integer}) `

`=>C`

Trigonometry, 2ADV T2 2019 HSC 13a

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

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`x = 0, quad pi/3, quad pi, quad (5 pi)/3`

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♦ Mean mark 49%.

`2 sin x cos x – sin x` `= 0`
`sin x (2 cos x – 1)` `= 0`
`sin x` `= 0`
`=> x` `= 0,\  pi,\  2pi`
`cos x` `= 1/2`
`=> x` `= pi/3, (5 pi)/3`

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

Trigonometry, 2ADV T2 SM-Bank 40

Let  `(tantheta - 1) (sin theta - sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`.

  1. State all possible values of  `tan theta`.  (1 mark)

  2. Hence, find all possible solutions for  `(tan theta - 1) (sin^2 theta - 3 cos^2 theta) = 0`, where  `0 <= theta <= pi`.  (2 marks)

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  1. `tan theta = 1 or tan theta = +- sqrt 3`
  2. `theta = pi/4, pi/3 or (2 pi)/3`
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i.  `(tantheta – 1) (sin theta – sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`

`=> tan theta = 1`

♦ Mean mark 42%.
`=>sin theta – sqrt 3 cos theta` `=0`
`sin theta` `=sqrt3 cos theta`
`tan theta` `=sqrt3`

 

`=>sin theta + sqrt 3 cos theta` `=0`
`sin theta` `=-sqrt3 cos theta`
`tan theta` `=-sqrt3`

 
`:. tan theta = 1 or tan theta = +- sqrt 3`

 

ii.  `(tan theta – 1) (sin^2 theta – 3 cos^2 theta) = 0`

`text(Using part a:)`

♦ Mean mark 42%.

`(tan theta – 1) (sin theta – sqrt 3 cos theta) (sin theta + sqrt 3 cos theta) = 0`

`=> tan theta` `= 1` `qquad or qquad` `tan theta` `= +- sqrt 3`
`theta` `= pi/4`   `theta` `= pi/3, (2 pi)/3`

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

Trigonometry, 2ADV T2 SM-Bank 34

Solve the equation  `sqrt 3 sin x = cos x`  for  `– pi<=x<= pi`.  (2 marks)

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`x = pi/6,\ \ \ – (5 pi)/6`

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`text(Divide both sides by)\ \ cos x :`

MARKER’S COMMENT: Many students who found the base angle correctly could not solve within the restrictions.
`sqrt 3 sin x` `=cos x`
`sqrt 3 tan x` `= 1`
`tan x` `= 1/sqrt 3`
`=>\ text(Base angle)\ = pi/6`

 
`:. x = pi/6\ \ text(or)\ – (5 pi)/6,\ \ \ (– pi <=x<= pi)`

Trigonometry, 2ADV T2 SM-Bank 32

Express  `5cot^2 x - 2text(cosec)\ x + 2`  in terms of  `text(cosec)\ x`  and hence solve

`5cot^2 x - 2text(cosec)\ x + 2 = 0`  for  `0 < x < 2pi`.  (3 marks)

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`x = pi/2`

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`cot^2 x` `= (cos^2 x)/(sin^2 x)`
  `= (1 – sin^2 x)/(sin^2 x)`
  `= text(cosec)^2 x – 1`

 

`5cot^2 x – 2text(cosec)\ x + 2` `= 0`
`5(text(cosec)^2 x – 1) – 2text(cosec)\ x + 2` `= 0`
`5text(cosec)^2 x – 2text(cosec)\ x – 3` `= 0`
`(5text(cosec)\ x + 3)(text(cosec)\ x – 1)` `= 0`
`text(cosec)\ x` `= −3/5` `text(cosec)\ x` `= 1`
`sinx` `= −5/3` `sinx` `= 1`
`(text(no solution))` `x` `= pi/2`

 
`:. x = pi/2`

Trigonometry, 2ADV T2 2016 HSC 8 MC

How many solutions does the equation  `|\ cos (2x)\ | = 1`  have for  `0 <= x <= 2 pi?`

  1. `1`
  2. `3`
  3. `4`
  4. `5` 
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`D`

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`|\ cos (2x)\ | = 1`

♦♦♦ Mean mark 23%.

`cos (2x) = +- 1`

`text(When)\ \ cos (2x)` `= 1`
`2x` `= 0, 2pi, 4 pi, …`
`:. x` `= 0, pi, 2 pi, …`

 

`text(When)\ \ cos (2x)` `= – 1`
`2x` `= pi, 3 pi, 5 pi, …`
`:. x` `= pi/2, (3 pi)/2, (5 pi)/2, …`

 

`:. x = 0, pi/2, pi, (3 pi)/2, 2 pi\ \ \ text(for)\ \ \ 0 <= x <= 2pi`

`=>  D`

Trigonometry, 2ADV T2 2004 HSC 9a

Consider the geometric series  `1 − tan^2 theta + tan^4 theta − …`

  1. When the limiting sum exists, find its value in simplest form.  (2 marks)

  2. For what values of  `theta`  in the interval
     
        `−pi/2 < theta < pi/2`  does the limiting sum of the series exist?  (2 marks)

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  1. `cos^2 theta`
  2. `− pi/4 < theta < pi/4`
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i.   `1 − tan^2 theta + tan^4 theta − …`

`=>\ text(GP where)\ \ a=1,\ \ r=T_2/T_1= − tan^2 theta`

`:. S_∞` `= 1/(1 − (−tan^2 theta))`
  `= 1/(1 + tan^2 theta)`
  `= 1/(sec^2 theta)`
  `= cos^2 theta`

 

ii.   `text(Find)\ \ theta\ \ text(such that)\ \ \ |\ r\ |` `< 1`
  `|−tan^2 theta\ |` `< 1`
  ` tan^2 theta` `< 1`
  `−1 < tan theta` `< 1`
  `:. − pi/4 < theta` `< pi/4`

Trigonometry, 2ADV T2 2004 HSC 8a

  1. Show that  `cos theta tan theta = sin theta`.  (1 mark)

  2. Hence solve  `8 sin theta cos theta tan theta = text(cosec)\ theta`  for  `0 ≤ theta ≤ 2pi`.  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `pi/6, (5pi)/6`
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i.    `text(Prove)\ \ cos theta tan theta = sin theta`

`text(LHS)` `= cos theta tan theta`
  `= cos theta ((sin theta)/(cos theta))`
  `= sin theta`
  `=\ text{RHS}`

 

ii.    `8 sin theta cos theta tan theta` `= text(cosec)\ theta`
   `:. 8 sin theta(sin theta)` `= text(cosec)\ theta,\ \ \ \ text{(part (i))}` 
  `8 sin^2 theta`  `= 1/(sin theta)` 
  `8 sin^3 theta`  `= 1` 
  `sin^3 theta`  `= 1/8` 
  `sin theta`  `= 1/2` 
   `:. theta` `= pi/6, (5pi)/6.\ \ \ \ text{(for}\ \ 0 ≤ theta ≤ 2pi text{)}` 

Trigonometry, 2ADV T2 2014 HSC 15a

Find all solutions of  `2 sin^2 x + cos x − 2 = 0`, where  `0 <= x <= 2pi`.   (3 marks)

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

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♦ Mean mark 42%
`2 sin^2 x + cos x\ – 2` `= 0`
`2(1\ – cos^2x) + cos x\ – 2` `= 0`
`2\ – 2cos^2x + cosx\ – 2` `= 0`
`-2cos^2x + cosx` `= 0`
`cosx (-2 cosx + 1)` `= 0`

 

`:. -2 cosx + 1` `= 0` `\ text(or)\ \ \ \ \ \ \ ` `cos x` `= 0`
`2 cos x` `= 1`   `x` `= pi/2,\ (3pi)/2`
`cos x` `= 1/2`      
`cos(pi/3)` `=1/2`      

 

`text(S)text(ince cos is positive in)\ 1^text(st) // 4^text(th)\ text(quadrants,)`

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

 

`:. x = pi/3,\ pi/2,\ (3pi)/2,\ (5pi)/3\ \ text(for)\ \ 0 <= x <= 2pi`

Trigonometry, 2ADV T2 2014 HSC 7 MC

How many solutions of the equation  `(sin x - 1)(tan x + 2) = 0`  lie between  `0`  and  `2 pi`?

  1. `1`
  2. `2`
  3. `3`
  4. `4`
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`B`

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♦♦♦ Mean mark 25%, making it the toughest MC question in the 2014 exam.
COMMENT: Note that the “2 solutions” answer relies on the sum of an infinity of zeros not equalling zero. This concept created unintended difficulty in this question.

`text(When)\ (sin x\ – 1)(tan x + 2) = 0`

`(sinx\ – 1) = 0\ \ text(or)\ \ tan x + 2 = 0`

`text(If)\ \ sin x\ – 1` `= 0`
`sin x` `= 1`
`x` `= pi/2,\ \ \ 0 < x < 2 pi`
`text(If)\ \ tan x + 2` `= 0`
`tan x` `= -2`

 

`=>\ text(Note that since)\ \ tan\ pi/2\ \ text(is undefined, there)`

`text(are only 2 solutions when)\ \ tan x = -2`

`text{(which occurs in the 1st and 4th quadrants).}`

 

`:.\ 2\ text(solutions)`

`=>  B`

Calculus, 2ADV C4 2010 HSC 5b

  1. Prove that  `sec^2 x + secx tanx = (1 + sinx)/(cos^2x)`.   (1 mark)

  2. Hence prove that  `sec^2 x + secx tanx = 1/(1 - sinx)`.     (1 mark)

  3. Hence, use the identity  `int sec ax tan ax\ dx=1/a sec ax`  to find the exact value of

     

          `int_0^(pi/4) 1/(1 - sinx)\ dx`.   (2 marks)

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  1. `text(Proof)  text{(See Worked Solutions)}`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `sqrt2`
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i.    `text(Need to prove)`

`sec^2x + secxtanx = (1 + sinx)/(cos^2x)`
 

`text(LHS)` `=sec^2x + secx tanx`
  `=1/(cos^2x) + 1/(cosx) xx (sinx)/cosx`
  `=1/(cos^2x) + (sinx)/(cos^2x)`
  `=(1 + sinx)/(cos^2x)`
  `= text(RHS)\ \ \ \ text(… as required)`

 

ii.   `text(Need to prove)` 

♦♦ Mean mark 31%.
`sec^2x + secx tanx` `= 1/(1\ – sinx)`
`text(i.e.)\ \ (1 + sinx)/(cos^2x)` `= 1/(1\ – sin x)\ \ \ \ \ text{(part (i))}`
`text(LHS)` `= (1 + sinx)/(cos^2x)`
  `=(1 + sin x)/(1\ – sin^2x)`
  `=(1 + sinx)/((1\ – sinx)(1 + sinx)`
  `=1/(1\ – sinx)\ \ \ \ text(… as required)`

 

iii.  `int_0^(pi/4) 1/(1\ – sinx)\ dx`

♦ Mean mark 37%.

`= int_0^(pi/4) (sec^2x + secx tanx)\ dx`

`= [tanx + secx]_0^(pi/4)`

`= [(tan(pi/4) + sec(pi/4)) – (tan0 + sec0)]`

`= [(1 + 1/(cos(pi/4)))\ – (0 + 1/(cos0))]`

`= 1 + sqrt2\ – 1`

`= sqrt2`