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Trigonometry, EXT1 T1 2023 HSC 5 MC

Which of the following is the value of   \(\sin ^{-1}(\sin a)\)  given that  \(\pi<a<\dfrac{3 \pi}{2}\)? 

  1. \(a-\pi\)
  2. \(\pi-a\)
  3. \(a\)
  4. \(-a\)
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\(B\)

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\(\sin(a)\) \(=-\sin(a-\pi)\ \ \text{(see above)}\)  
\(\sin^{-1}(\sin(a))\) \(=\sin^{-1}(-\sin(a-\pi))\)  
  \(=\sin^{-1}(\sin(\pi-a)) \)  
  \(=\pi-a \)  

 
\(\Rightarrow B\)

Mean mark 56%.

Trigonometry, EXT1 T1 2022 HSC 1 MC

It is given that  `cos((23 pi)/(12))=(sqrt6+sqrt2)/(4)`.

Which of the following is the value of  `cos^(-1)((sqrt6+sqrt2)/(4))`?

  1. `{23pi}/{12}`
  2. `{11pi}/{12}`
  3. `pi/12`
  4. `- {11pi}/{12}`
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`C`

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`(23pi)/12\ \ =>\ \ text{4th quadrant}`

`cos((23pi)/12)=cos(2pi-(23pi)/12)=cos(pi/12)`

`:. cos^(-1)((sqrt6+sqrt2)/(4))=pi/12`

`=>C`


Mean mark 54%.

Trigonometry, EXT1 T1 SM-Bank 3

Determine the exact value of

`sin^(-1)(sqrt3/2)-sin^(-1)(-sqrt3/2)-cos^(-1)(-1/2)`.  (3 marks)

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`0`

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`text(Base angle:)\ \ sin\ pi/3 = sqrt3/2`

`sin =>\ text(– ve in 4th quadrant)`

`sin^(-1)(sqrt3/2)` `= pi/3`
`sin^(-1)(-sqrt3/2)` `= -pi/3`

 
`text(Base angle:)\ \ cos\ pi/3 = 1/2`

`cos =>\ text(– ve in 2nd quadrant)`

`cos^(-1)(-1/2)` `= pi-pi/3= (2pi)/3`

 
`:. sin^(-1)(sqrt3/2)-sin^(-1)(-sqrt3/2)-cos^(-1)(-1/2)`

`= pi/3-(-pi/3)-(2pi)/3`

`= 0`

Trigonometry, EXT1 T1 SM-Bank 1

Find `sin theta` given that  `theta = arccos (12/13) + arctan (3/4)`, and  `0<=theta<=pi/2`.   (3 marks)

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`56/65`

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`sin theta` `= sin (cos^(-1) (12/13) + tan^(-1)(3/4))`
  `= sin (cos^(-1) (12/13)) cos (tan^(-1) (3/4)) + sin (tan^(-1)(3/4)) cos (cos^(-1)(12/13))`

 
 

 

COMMENT: Syllabus states that students must understand and use the notation `arccos,\ arcsin` etc..

`:. sin theta` `= 5/13 xx 4/5 + 3/5 xx 12/13`
  `= 20/65 + 36/65`
  `= 56/65`

Trigonometry, EXT1 T1 2007 HSC 5c

Find the exact values of `x` and `y` which satisfy the simultaneous equations

`sin^(-1)\ x + 1/2\ cos^(-1)\ y = pi/3`   and

`3\ sin^(-1)\ x-1/2\ cos^(-1)\ y = (2pi)/3`. (3 marks)

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`x = 1/sqrt2, \ \ \ y = sqrt3/2`

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`sin^(-1)\ x + 1/2\ cos^(-1)\ y` `= pi/3` `\ \ …\ (1)`
`3\ sin^(-1)\ x-1/2\ cos^(-1)\ y` `= (2pi)/3` `\ \ …\ (2)`

 

MARKER’S COMMENT: The use of ther elimination method here proved much more successful than substitution.

`text(Add)\ \ (1) + (2)`

`4\ sin^(-1)\ x` `= pi`
`sin^(-1)\ x` `= pi/4`
`:.x` `= 1/sqrt2`

 

`text(Substitute)\ \ x = 1/sqrt2\ \ text(into)\ (1)`

`sin^(-1)\ 1/sqrt2 + 1/2\ cos^(-1)\ y` `= pi/3`
`pi/4 + 1/2\ cos^(-1)\ y` `= pi/3`
`1/2\ cos^(-1)\ y` `= pi/12`
`cos^(-1)\ y` `= pi/6`
`:.y` `= sqrt3/2`

Trigonometry, EXT1 T1 2012 HSC 13a

Write  `sin(2 cos ^(-1) (2/3))`  in the form  `a sqrtb`, where `a` and `b` are rational.   (2 mark)

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 `4/9 sqrt5` 

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TIP: This question is made  less complicated by reminding yourself that `cos^(-1) (2/3)` is simply an angle.

`sin (2 cos^(-1) (2/3))`

`text(Let)\ \ x = cos ^(-1) (2/3)`

`:.\ cos x = 2/3`
 

Inverse Functions, EXT1 2012 HSC 13a Answer 
 

`sin x = sqrt5/3`

`sin(2 cos^(-1) (2/3))` `= sin 2x`
  `= 2 sin x cos x`
  `= 2 xx sqrt5/3 xx 2/3`
  `= 4/9 sqrt5`