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Trigonometry, EXT1 T3 2025 HSC 12e

  1. Express  \(\sqrt{3} \, \sin x-\cos x\)  in the form  \(2\, \sin (x-\alpha)\), where  \(0<\alpha<\dfrac{\pi}{2}\).   (1 mark)

  2. Hence, or otherwise, solve  \(\sqrt{3}\, \sin x=\cos x+1\),  where  \(0 \leq x \leq 2 \pi\).   (2 marks)

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i.    \(\text{See Worked Solutions}\)

ii.   \(x=\dfrac{\pi}{3}, \pi\)

Show Worked Solution

i.     \(2 \sin (x-\alpha)=\sqrt{3} \sin x-\cos x\)

\(2 \sin x \, \cos \alpha-2 \cos x \, \sin \alpha=\sqrt{3} \sin x-\cos x\)

\(\text{Equating coefficients:}\)

\(2 \cos \alpha=\sqrt{3}, \ \ 2 \sin \alpha=1\)

\(\text{Since} \ \ \cos \alpha>0 \ \ \text{and} \ \ \sin \alpha>0\)

\(\Rightarrow \alpha \ \text{is in 1st quadrant.}\)

\(\tan \alpha=\dfrac{1}{\sqrt{3}} \ \Rightarrow \ \alpha=\dfrac{\pi}{6}\)

\(\therefore \sqrt{3}\, \sin x-\cos x=2\, \sin \left(x-\dfrac{\pi}{6}\right)\)
 

ii.     \(\sqrt{3} \sin x\) \(=\cos x+1\)
  \(\sqrt{3} \sin x-\cos x\) \(=1\)
  \(2 \sin \left(x-\dfrac{\pi}{6}\right)\) \(=1\)
  \(\sin \left(x-\dfrac{\pi}{6}\right)\) \(=\dfrac{1}{2}\)
  \(x-\dfrac{\pi}{6}\) \(=\dfrac{\pi}{6}, \dfrac{5 \pi}{6}\)
  \(x\) \(=\dfrac{\pi}{3}, \pi \quad(0 \leqslant x \leqslant 2 \pi)\)

Trigonometry, EXT1 T3 EQ-Bank 2

The tides in Ulladulla Harbour can be modelled to the equation

\(h=2 \sqrt{3} \cos (2 t)+2 \sin (2 t)\)

where \(h\) is the height of the tide in metres and \(t\) is the time in hours after midnight.

Large fishing vessels can only enter or leave the harbour when the tide is at least 2 meters high.

Determine the first time after midnight that large shipping vessels can no longer enter or exit the harbour and when access is again possible, giving your answers to the nearest minute?   (4 marks)

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\(\text{Vessels can not enter/exit the harbour between 00:47 and 02:53.}\)

Show Worked Solution

\(A \sin (x+2 t)=2 \sqrt{3} \cos (2 t)+2 \sin (2 t) \)

\(A \sin x \cos (2 t)+A \cos x \sin (2 t)=2 \sqrt{3} \cos (2 t)+2 \sin (2 t)\)

\(\text{Equating coefficients:}\)

\(A \sin x=2 \sqrt{3}\)

\(A \cos x=2 \)

\(A^2=(2 \sqrt{3})^2+2^2=16 \ \Rightarrow \ \ A=4 \)
 

\(\dfrac{A \sin x}{A \cos x}=\dfrac{2 \sqrt{3}}{2} \ \Rightarrow \ \tan x=\sqrt{3} \ \Rightarrow \ x=\dfrac{\pi}{3} \)

\(\Rightarrow \ h=4 \sin \left(\dfrac{\pi}{3}+2 t\right)\)
 

\(\text{Find \(t\) when  \(h=2\):}\)

\(2=4 \sin \left(\dfrac{\pi}{3}+2 t\right)\)

\(\dfrac{\pi}{3}+2 t=\sin ^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{6}, \dfrac{5 \pi}{6}, \dfrac{13 \pi}{6}, \cdots\)

\begin{array}{l|l}
2 t_1=\dfrac{5 \pi}{6}-\dfrac{\pi}{3}=\dfrac{\pi}{2} \quad & \quad2 t_2=\dfrac{13 \pi}{6}-\dfrac{\pi}{3}=\dfrac{11 \pi}{6}\\
t_1=\dfrac{\pi}{4} & \quad t_2=\dfrac{11 \pi}{12}\\
\ =0.7853 \ldots \text{hours} & \quad\ \ \ =2.879 \ldots \text {hours}\\
\ \ = \ \text{47 mins (after 12)} & \quad \quad=\ \text{2 hr 53 mins}\\
\end{array}

\(\therefore\ \text{Vessels can not enter/exit the harbour between 00:47 and 02:53.}\)

Trigonometry, EXT1 T3 2024 HSC 10 MC

For real numbers \(a\) and \(b\), where  \(a \neq 0\)  and  \(b \neq 0\), we can find numbers \(\alpha\), \(\beta\), \(\gamma\), \(\delta\) and \(R\) such that  \(a\,\cos x + b\,\sin x\)  can be written in the following 4 forms:

\(R\,\sin(x + \alpha)\)

\(R\,\sin(x-\beta)\)

\(R\,\cos(x + \gamma)\)

\(R\,\cos(x-\delta)\)

where  \(R \gt 0\)  and  \(0<\alpha, \beta, \gamma, \delta \lt 2\pi\).

What is the value of  \(\alpha + \beta + \gamma + \delta\)?

  1. \(0\)
  2. \(\pi\)
  3. \(2\pi\)
  4. \(4\pi\)
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\(D\)

Show Worked Solution

\(a\,\sin\,x+b\,\cos\,x\ \ \text{can be written 4 ways.}\)

\(\text{Consider the case:}\)

\(\cos\,x+\sin\,x\ \ \text{where}\ a=b=1,\ \ R=\sqrt{2}\)

\(\sqrt{2}\,\sin(x+\alpha)\) \(= \sqrt{2}(\sin\,x\,\cos\,\alpha + \cos\,x\,\sin\,\alpha)\)  
  \(= \sqrt{2}\Big(\sin\,x \cdot \dfrac{1}{\sqrt2} + \cos\,x \cdot \dfrac{1}{\sqrt2}\Big)\)  

 
\(\cos\, \alpha^{+}, \sin\,\alpha^{+}\ \Rightarrow\ \ \alpha=\dfrac{\pi}{4} \)

♦♦♦ Mean mark 19%.

\(\text{Similarly for other 3 cases:}\)

\(\sqrt{2}\,\sin(x-\beta): \ \cos\, \beta^{+}, \sin\,\beta^{-}\ \Rightarrow\ \ \beta=\dfrac{7\pi}{4} \)

\(\sqrt{2}\,\cos(x+\gamma): \ \cos\, \gamma^{+}, \sin\,\gamma^{-}\ \Rightarrow\ \ \gamma=\dfrac{7\pi}{4} \)

\(\sqrt{2}\,\cos(x-\delta): \ \cos\, \delta^{+}, \sin\,\delta^{+}\ \Rightarrow\ \ \delta=\dfrac{\pi}{4} \)

\(\therefore \alpha + \beta + \gamma + \delta = 4\pi\)

\(\Rightarrow D\)

Trigonometry, EXT1 T3 2022 HSC 11e

Express  `sqrt3sin(x)-3cos(x)`  in the form  `R sin(x+alpha)`.  (3 marks)

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`2sqrt3 sin(x-pi/3)`

Show Worked Solution

`R sin(x+alpha)=sqrt3sin(x)-3cos(x)`

`R sin(x)cos alpha +Rcos(x)sin alpha=sqrt3sin(x)-3cos(x)` 

`text{Equating coefficients:}`

`R cos alpha=sqrt3, \ \ R sin alpha=-3`

`R^2` `=(sqrt3)^2+(-3)^2=12`  
`:.R` `=sqrt12=2sqrt3`  

 
`text{Given}\ \ sin alpha<0\ \ text{and}\ \ cos alpha>0`

`=> alpha\ \ text{is in the 4th quadrant}`

`sqrt12 cos alpha` `=sqrt3`  
`cos alpha` `=sqrt3/sqrt12=1/2`  
`:.alpha` `=-pi/3\ \ (-pi<=alpha<=pi)`  

 
`:.sqrt3sin(x)-3cos(x)=2sqrt3 sin(x-pi/3)`

Trigonometry, EXT1 T3 2020 HSC 11d

By expressing  `sqrt3 sin x + 3 cos x`  in the form  `A sin (x + a)`, solve  `sqrt3 sin x + 3 cos x = sqrt3`, for  `0 <= x <= 2pi`.  (4 marks)

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`x = pi/2, \ (11pi)/6`

Show Worked Solution

`A sin(x + alpha) = sqrt3 sinx + 3cosx`

`Asinx cos alpha + Acosx sin alpha = sqrt3 sinx + 3cosx`
 

`text(Equating co-efficients:)`

`=> Acos alpha = sqrt3`

`=> Asin alpha = 3`

`A^2` `= (sqrt3)^2 + 3^2 = 12`
`:. A` `= sqrt12`
`(Asinalpha)/(Acosalpha)` `= 3/sqrt3`
`tanalpha` `= sqrt3`
`alpha` `= pi/3`

 

`sqrt3 sinx + 3cosx` `= sqrt3`
`sqrt12 sin(x + pi/3)` `= sqrt3`
`sin(x + pi/3)` `= 1/2`
`x + pi/3` `= (5pi)/6, (13pi)/6`
`:. x` `= pi/2, (11pi)/6\ \ \ (0 <= x <= 2pi)`

Trigonometry, EXT1 T3 EQ-Bank 5

A particular energy wave can be modelled by the function

`f(t) = sqrt5 sin 0.2t + 2 cos 0.2t, \ \ t ∈ [0, 50]`

  1. Express this function in the form  `f(t) = Rsin(nt-alpha), \ \ alpha ∈ [0, 2pi]`.  (2 marks)

  2. Find the time the wave first attains its maximum value. Give your answer to one decimal place.  (2 marks)

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  1. `f(t) = 3sin(0.2t-5.553)`
  2. `t = 4.2`
Show Worked Solution

i.   `f(t) = sqrt5 sin 0.2t + 2cos 0.2t`

`Rsin(nt-alpha)` `= Rsin(0.2t-alpha)`
  `= Rsin 0.2t cosalpha-Rcos 0.2t sinalpha`

 

`=> Rcosalpha = sqrt5,\ \ R sinalpha = −2`

`R^2` `= (sqrt5)^2 + (-2)^2 = 9`
`R` `= 3`

 
`cosalpha = sqrt5/3, sinalpha = −2/3`

`=> alpha\ text(is in 4th quadrant)`
 

`text(Base angle) = cos^(−1)(sqrt5/3) = 0.7297`

`:.alpha` `= 2pi-0.7297`
  `= 5.553…`

 
`:. f(t) = 3sin(0.2t-5.553)`

 

ii.   `text(Max value occurs when)\ sin(0.2t-5.553) = 1`

`0.2t-5.553` `= pi/2`
`0.2t` `= 7.124…`
`t` `= 35.62…`

 
`text(Test if)\ \ t > 0\ \ text(for:)`

`0.2t-5.553` `= -(3pi)/2`
`0.2t` `= 0.8406`
`t` `= 4.20…`

 
`:. text(Time of 1st maximum:)\ \ t = 4.2`

Trigonometry, EXT1 T3 EQ-Bank 4

The current flowing through an electrical circuit can be modelled by the function

`qquad f(t) = 6sin 0.05t + 8cos 0.05t, \ \ t >= 0`

  1.  Express the function in the form  `f(t) = Asin(at + b),\ \ text(for)\ \ 0<=b<=pi/2`.  (2 marks)

  2.  Find the time at which the current first obtains it maximum value.   (1 mark)

  3.  Sketch the graph of  `f(t)`. Clearly show its range and label the coordinates of its first maximum value. Do not label `x`-intercepts.   (1 mark)

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  1. `f(t) = 10sin(0.05t + 0.927)`
  2.   `12.9\ \ (text(to 1 d.p.))`
  3.  
Show Worked Solution

i.   `f(t) = 6sin 0.05t + 8cos 0.05t`

`Asin(at + b)` `=Asin(0.05t + b)`  
  `= Asin(0.05t)\ cos(b) + Acos(0.05t)\ sin(b)`  

 
`=> Acos(b) = 6, \ Asin(b) = 8`

`A^2` `= 6^2 + 8^2`
`A` `= 10`

 

`=> 10cos(b)` `= 6`
`cos(b)` `= 6/10`
`b` `= cos^(−1) 0.06 ~~ 0.927\ text(radians)`

 
`:. f(t) = 10sin(0.05t + 0.927)`

 

ii.   `text(Max occurs at)\ \ sin(0.05t + 0.927)= sin\ pi/2`

`0.05t + 0.927` `= pi/2`
`0.05t` `= 0.643…`
`:.t` `= 12.87`
  `= 12.9\ \ (text(to 1 d.p.))`

 

iii.   

Trigonometry, EXT1 T3 2018 HSC 11c

Write  `sqrt 3 sin x + cos x`  in the form  `R sin (x + alpha)`  where  `R > 0`  and  `0 <= alpha <= pi/2.`  (2 marks)

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`text(See Worked Solution)`

Show Worked Solution
`R sin (x + alpha)` `=sqrt 3 sin x + cos x`
`R sin x cos alpha + R cos x sin alpha` `=sqrt 3 sin x + cos x`

 
`text(Equating coefficients:)`

`=> R cos alpha` `= sqrt 3`  
`=> R sin alpha` `= 1`  

 

`R^2` `= (sqrt 3)^2 + 1^2=4`
`:.R` `= 2`

 

`=> 2sin alpha` `=1`
`sin alpha` `= 1/2`
`alpha` `= pi/6 qquad (0 <= alpha <= pi/2)`

 
`:. sqrt 3 sin x + cos x = 2 sin (x + pi/6).`

Trigonometry, EXT1 T3 2017 HSC 4 MC

What is the value of  `tan alpha`  when the expression  `2sinx - cosx`  is written in the form  `sqrt5 sin(x - alpha)`?

A.     `−2`

B.     `−1/2`

C.     `1/2`

D.     `2`

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`C`

Show Worked Solution

`text(Using)\ \ sqrt5 sin(x – alpha) = sqrt5 sinx cosalpha – sqrt5cosxsinalpha,`

`=>sqrt5 sinx cosalpha – sqrt5 cosx sinalpha = 2sinx – cosx`

`=> sqrt5 cosalpha = 2, sqrt5 sinalpha = 1`

`(sinalpha)/(cosalpha)` `= 1/2`
`:. tan alpha` `= 1/2`

`⇒C`

Trigonometry, EXT1 T3 2009 HSC 2b

  1. Express  `3 sin x + 4 cos x`  in the form  `A sin(x + alpha)`  where  `0 <= alpha <= pi/2`.  (2 marks)

  2. Hence, or otherwise, solve  `3 sin x + 4 cos x = 5`  for  `0 <= x <= 2pi`.

     

    Give your answer, or answers, correct to two decimal places.    (2 marks)

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  1. `5 sin (x + cos^-1 (3/5))\ \ text(or)`

     

    `5 sin (x+0.93)`

  2. `0.64\ text(radians)\ text{(2 d.p.)}`
Show Worked Solution
i.    `A sin (x + alpha) = 3 sin x + 4 cos x`
  `A sinx cos alpha + A cos x sin alpha = 3 sin x + 4 cos x`

 

`=> A cos alpha = 3\ \ \ \ \ A sin alpha = 4`

`A^2` `= 3^2 + 4^2 = 25`
`:. A` `= 5`
`=> 5 cos alpha` `= 3`
`cos alpha` `= 3/5`
`alpha` `= cos^(-1) (3/5)= 0.9272… \ \ text(radians)`

 

`:.\ 3 sin x + 4 cos x = 5 sin (x + cos^-1 (3/5)) `

 

ii.   `3 sin x + 4 cos x` `= 5`
  `5 sin (x + alpha)` `= 5`
  `sin (x + alpha)` `= 1`
  `x + alpha` `= pi/2, (5pi)/2, …`
  `x` `= pi/2\ – 0.9272…\ \ \ \ \ (0 <= x <= 2 pi)`
    `= 0.6435…`
    `= 0.64\ text(radians)\ text{(2 d.p.)}`