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Trigonometry, EXT1 T3 2021 HSC 11g

By factorising, or otherwise, solve  `2sin^3x + 2sin^2x - sinx - 1 = 0`  for  `0 <= x <= 2pi`.  (3 marks)

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`x = pi/4, (3pi)/4, (5pi)/4, (3pi)/2, (7pi)/4`

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`2sin^3x + 2sin^2x – sinx – 1 = 0`

`2sin^2x (sinx + 1) – (sinx + 1)` `= 0`
`(2sin^2x – 1)(sinx + 1)` `= 0`
`2sin^2 x` `= 1`    `sinx =` `= -1`
`sin^2x` `= 1/2`    
`sinx` `= ± 1/sqrt2`    

 
`:. x = pi/4, (3pi)/4, (5pi)/4, (3pi)/2, (7pi)/4`

Trigonometry, EXT1 T3 SM-Bank 11

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 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`

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`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 SM-Bank 1

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)\ \ text{(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)`

 

`text(S)text(ince)\ \ theta` `= A – B`
`theta` `= 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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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-1/18 cos(9x)-1/2 cos(x) + c`
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i.   `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)`

 

ii.  `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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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
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i.   `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)`

 

ii.   `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`