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Trigonometry, EXT1 T3 2024 HSC 14c

  1. Explain why the equation  \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta\), where  \(-\pi<\theta<\pi\), has exactly one solution.   (1 marks)

  2. Solve  \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}\).   (2 marks)

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i.     \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi\)

\(\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)

 \(\Rightarrow \text { Both are monotonically increasing functions}\)

\(\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)\) 

 \(\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}\)
 

ii.   \(x=\dfrac{1}{2}\)

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i.     \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi\)

\(\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)

♦♦♦ Mean mark (i) 11%.

\(\Rightarrow \text { Both are monotonically increasing functions}\)

\(\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)\) 

 \(\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}\)
 

ii.    \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}\)

\(\tan \left(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)\right)=\tan \left(\dfrac{3 \pi}{4}\right)\)

♦♦ Mean mark (ii) 32%.

\(\dfrac{\tan \left(\tan ^{-1}(3 x)\right)+\tan \left(\tan ^{-1}(10 x)\right)}{1-\tan \left(\tan ^{-1}(3 x)\right) \cdot \tan \left(\tan ^{-1}(10 x)\right)}=-1\)

  \(\dfrac{3 x+10 x}{1-30 x^2}\) \(=-1\)
  \(13 x\) \(=30 x^2-1\)
  \(30 x^2-13 x-1\) \(=0\)
  \((15 x+1)(2 x-1)\) \(=0\)

 
\(x=\dfrac{1}{2}\ \ \text {or}\ \ -\dfrac{1}{15}\)

\(\text {Graph is monotonically increasing through } (0,0) \Rightarrow \ \Big(x \neq -\dfrac{1}{15} \Big)\)

\(\therefore x=\dfrac{1}{2}\)

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)`
    `= 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 28

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

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

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a.    `text(See Worked Solutions)`

b.    `x = 0, pi/2, pi, (3pi)/2, 2pi, (2pi)/3, (4pi)/3`

Show Worked Solution
a.   `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)`

 

b.    `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 26

Show that

`cos3x = 4cos^3 x-3cosx`.   (3 marks)

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`text(See Worked Solutions)`

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COMMENT: 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 EQ-Bank 24

Given that  `cos (theta-phi) = 3/5`  and  `tan theta tan phi = 2`, find  `cos(theta + phi)`.   (3 marks)

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`-1/5`

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`cos(theta+phi)= cos theta cos phi-sin theta sin phi`

`cos theta cos phi + sin theta sin phi = 3/5\ \ …\ (1)`

`(sin theta sin phi)/(cos theta cos phi)` `=2`  
`sin theta sin phi` `= 2 cos theta cos phi\ \ …\ (2)`  

  
`text{Substitute (2) into (1):}`

`cos theta cos phi + 2 cos theta cos phi` `= 3/5`
`3 cos theta cos phi` `= 3/5`
`cos theta cos phi` `= 1/5`

 
`text(Substitute)\ \ cos theta cos phi=1/5\ \ text{into (1):}`

`1/5 + sin theta sin phi` `= 3/5`
`sin theta sin phi` `= 2/5`

 

`:. cos(theta + phi)` `= cos theta cos phi-sin theta sin phi`
  `= 1/5-2/5`
  `= -1/5`

Trigonometry, EXT1 T3 EQ-Bank 30

A billboard of height `a` metres is mounted on the side of a building, with its bottom edge `h` metres above street level. The billboard subtends an angle `theta` at the point `P`, `x` metres from the building.
 

 
Use the identity  `tan (A-B) = (tan A-tan B)/(1 + tanA tanB)`  to show that

`theta = tan^(-1) [(ax)/(x^2 + h(a + h))]`.   (2 marks)

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`text{Proof (See Worked Solutions)}`

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`text(Consider angles)\ \ A and B\ \ text(on the graph:)`

MARKER’S COMMENT: Answers that included a diagram and clearly labelled angles were generally successful.

`text(Show)\ \ theta = tan^(-1) [(ax)/(x^2 + h(a + h))]`

`tan A= (a + h)/x, \ tan B= h/x`

`tan (A-B)` `= ((a + h)/x-h/x)/(1 + ((a + h)/x)(h/x)) xx (x^2)/(x^2)`
  `= (x(a + h)-xh)/(x^2 + h(a + h))`
  `= (ax)/(x^2 + h(a + h)`
`A-B` `= tan^(-1) [(ax)/(x^2 + h(a + h))]\ \ \ text(… as required.)`

Calculus, EXT1 C2 2005 HSC 3b

  1. By expanding the left-hand side, show that
  2. `qquad sin(5x + 4x) + sin(5x-4x) = 2 sin (5x) cos(4x)`   (1 mark)

  3. Hence find  `int sin(5x) cos (4x)\ dx.`   (2 marks)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `-1/18 cos(9x)-1/2 cos(x) + c`

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a.    `sin (5x + 4x) + sin (5x-4x) = 2 sin(5x) cos(4x)`

`text(LHS)` `= sin (5x) cos (4x)-sin(4x) cos (5x) + sin (5x) cos (4x)+ sin (4x) cos (5x)`
  `= 2 sin (5x) cos (4x)\ \ text(…  as required)`

 

b.  `int sin (5x) cos (4x)\ dx`

`= 1/2 int 2 sin (5x) cos (4x)\ dx`

`= 1/2 int sin (5x + 4x) + sin (5x-4x)\ dx`

`= 1/2 int sin (9x) + sin (x)\ dx`

`= 1/2 [-1/9 cos(9x)-cos(x)] + c`

`= -1/18 cos(9x)-1/2 cos(x) + c`

Trigonometry, EXT1 T3 2008 HSC 6b

It can be shown that  `sin 3 theta = 3 sin theta-4 sin^3 theta`  for all values of  `theta`. (Do NOT prove this.)

Use this result to solve  `sin 3 theta + sin 2 theta = sin theta`  for  `0 <= theta <= 2pi`.   (3 marks)

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`theta = 0, pi/3, pi, (5pi)/3, 2pi\ \ \ text(for)\ \ \ 0 <= theta <= 2 pi`

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`text(Substitute)\ \ sin 3 theta = 3 sin theta-4 sin^3 theta`

`text(into)\ \ sin 3 theta + sin 2 theta = sin theta,`

`3 sin theta-4 sin^3 theta + sin 2 theta` `= sin theta`
`2 sin theta-4 sin^3 theta + 2 sin theta cos theta` `= 0`
`2 sin theta [1-2 sin^2 theta + cos theta]` `= 0`
`2 sin theta [1-2(1-cos^2 theta) + cos theta]` `= 0`
`2 sin theta [ 1-2 + 2 cos^2 theta + cos theta]` `= 0`
`2 sin theta [2 cos^2 theta + cos theta-1]` `= 0`
`2 sin theta (2 cos theta-1)(cos theta + 1)` `= 0`

 
`2 sin theta = 0\ \ =>\ \ theta = 0, pi, 2pi`

`2 cos theta-1= 0\ \ =>\ \ cos theta= 1/2\ \ =>\ \ theta = pi/3, (5pi)/3`

`cos theta + 1= 0\ \ =>\ \ cos theta= -1\ \ =>\ \ theta=pi` 

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

Trigonometry, EXT1 T3 2010 HSC 6a

  1. Show that  `cos(A-B) = cos A cos B(1 + tan A tan B)`.   (1 mark)

  2. Suppose that  `0 < B < pi/2`  and  `B < A < pi`.     
  3. Deduce that if  `tan Atan B = − 1`, then  `A\-B = pi/2`.   (1 mark)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

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a.    `text(Show)\ \ cos(A-B) = cosA cosB (1 + tanA tanB)`

`text(RHS)` `= cosA cosB (1 + (sinA sinB)/(cosA cosB))`
  `= cosA cos B + sinA sin B`
  `= cos(A-B)\ text(… as required)`

 

b.    `text(Given)\ \ tanA tanB = -1`

`cos (A-B)` `= cosA cosB (1-1)`
`cos (A-B)` `= 0`
`A-B` `= cos^(-1) 0`
  `= pi/2, (3pi)/2, …`

 
`text(S)text(ince)\ \ \ 0 < B < pi/2\ \ text(and)\ \ \ B < A < pi,`

`=> A-B = pi/2`