Exact Trig Ratios (Adv-2027)

Trigonometry, 2ADV T2 2025 HSC 31

The equation \(\cos \, p x=\dfrac{1}{2}\) has 2 solutions where \(0 \leq x \leq 2 \pi\) and \(p>0\).

Find all possible values of \(p\). (3 marks)

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\(\dfrac{5}{6} \leqslant p<\dfrac{7}{6}\)

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\(\cos (p x)=\dfrac{1}{2}\ \ \Rightarrow\ \ px=\cos ^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{3}, \dfrac{5 \pi}{3}, \dfrac{7 \pi}{3}\)

\(\text{Since there are 2 solutions in range} \ \ 0 \leq x<2 \pi:\)

\(p x=\dfrac{5 \pi}{3} \ \Rightarrow \ x=\dfrac{5 \pi}{3 p}\)

\(\dfrac{5 \pi}{3 p}\)	\(\leqslant 2 \pi\)
\(3 p\)	\(\geqslant \dfrac{5}{2}\)
\(p\)	\(\geqslant \dfrac{5}{6}\)

\(\text{Since}\ \ p x=\dfrac{7 \pi}{3} \ \ \text{cannot be a solution in the range} \ \ 0 \leq x \leq 2 \pi:\)

\(p x=\dfrac{7 \pi}{3} \ \Rightarrow \ x=\dfrac{7 \pi}{3 p}\)

\(\dfrac{7 \pi}{3 p}\)	\(>2 \pi\)
\(p\)	\(<\dfrac{7}{6}\)

\(\therefore \dfrac{5}{6} \leqslant p<\dfrac{7}{6}\)

Trigonometry, 2ADV T2 EQ-Bank 1 MC

Determine the number of values of \(\theta\) in the range \(0^{\circ} \leqslant \theta \leqslant 360^{\circ}\) that satisfy the equation

\((\tan \theta-\sqrt{3})(\cos^{2}\theta-1)=0 \)

\(3\)
\(4\)
\(5\)
\(6\)

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\(C\)

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\(\tan \theta = \sqrt{3}\ \rightarrow \text{2 solutions} \)

\(\cos^{2}\theta\)	\(=1\)
\(\cos\theta\)	\(= \pm 1\)
\(\theta\)	\(=0^{\circ}, 180^{\circ}, 360^{\circ}\ \rightarrow \text{3 solutions} \)

\(\Rightarrow C\)

Trigonometry, 2ADV T2 2023 HSC 20

Find all the values of `theta`, where `0^@ <=theta <= 360^@`, such that

`sin(theta-60^@)=-sqrt3/2` (3 marks)

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`theta=0^@, 300^@ and 360^@`

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`sin60^@=sqrt3/2\ \ =>\ \ text{Base angle}\ =60^@`

`=>\ text{sin is negative in 3rd and 4th quadrants}`

`sin(theta-60^@)`	`=180+60, 360-60`
	`=240^@, 300^@`

`theta-60^@=240^@\ \ =>\ \ theta=300^@`

`theta-60^@=300^@\ \ =>\ \ theta=360^@`

`text{Consider}\ theta = 0^@`

`sin(0-60^@)=sin(-60^@)=-sqrt3/2`

`:.theta=0^@, 300^@ and 360^@`

Trigonometry, 2ADV T2 SM-Bank 43

Find the exact value of

`cot(-(5pi)/6)`. (2 marks)

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`sqrt3`

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`cot(-(5pi)/6)`	`= 1/(tan(-(5pi)/6))`
	`= 1/tan(pi/6)`
	`= 1/(1/sqrt3)`
	`= sqrt3`

Trigonometry, 2ADV T2 SM-Bank 39

Let `f(x) = sin((2pix)/3)`.

Solve the equation `sin((2pix)/3) = -sqrt3/2` for `0<=x<=3` (2 marks)

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`x = 2, 5/2`

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`sin((2pix)/3) = -sqrt3/2`

`=>\ text(Base angle)\ = pi/3`

`(2 pi x)/3`	`=(4pi)/3, (5pi)/3, (10pi)/3, …`
`:.x`	`=2 or 5/2, \ \ \ (0<=x<=3)`

Trigonometry, 2ADV T2 SM-Bank 38

Solve `2cos(2x) = −sqrt3` for `x`, where `0 <= x <= pi`. (2 marks)

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`x = (5pi)/12, (7pi)/12`

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`cos(2x)`	`= – sqrt3/2`
`2x`	`= (5pi)/6, 2pi – (5pi)/6, 2pi+(5pi)/6`
	`=(5pi)/6, (7pi)/6, (17pi)/6,\ …`
`:. x`	`=(5pi)/12, (7pi)/12\ \ \ (0 <= x <= pi)`

Trigonometry, 2ADV T2 SM-Bank 37

Solve the equation `cos((3x)/2) = 1/2` for `−pi/2<=x<=pi/2`. (2 marks)

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`x = ± (2pi)/9`

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`cos((3x)/2) = 1/2`

`=>\ text(Base angle)\ =pi/3`

`(3x)/2`	`= (-pi)/3, pi/3, (5pi)/3, …`
`:. x`	`=(-2pi)/9, (2pi)/9, (10pi)/9`
	`= (-2pi)/9, (2pi)/9\ \ \ (-pi/2<=x<=pi/2)`

Trigonometry, 2ADV T2 SM-Bank 36

Solve the equation `sin (x/2) = -1/2` for `2 pi<=x<= 4 pi` (2 marks)

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`x = (7 pi)/3, (11 pi)/3`

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`x/2`	`=pi/6 + pi, 2pi-pi/6, 2pi + (pi/6 +pi), …`
	`=(7pi)/6, (11pi)/6, (19pi)/6, …`
`:. x`	`=(7pi)/3, (11pi)/3, (19pi)/3, …`

`text(Given)\ \ \2 pi<=x<= 4 pi`

`:. x = (7 pi)/3, (11 pi)/3`

Trigonometry, 2ADV T2 SM-Bank 35

Solve the equation

`sin (2x + pi/3) = 1/2\ \ text(for)\ \ 0<= x <=pi` (2 marks)

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`x = pi/4, (11 pi)/12`

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`sin (2x + pi/3) = 1/2`

`=>\ text(Base angle is)\ \ pi/6`

`(2x + pi/3)`	`= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, …`
`2x`	`= – pi/6, pi/2, (11pi)/6, (15pi)/6, …`
`x`	`= – pi/12, pi/4, (11pi)/12, (15pi)/12, …`

`:. x = pi/4, (11 pi)/12\ \ (0<=x<=pi)`

Trigonometry, 2ADV T2 2016 HSC 11g

Solve `sin (x/2) = 1/2` for `0 <= x <= 2pi.` (2 marks)

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

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`sin\ x/2 = 1/2\ \ text(for)\ \ 0 <= x <= 2pi`

`=>\ text(Reference angle) = pi/6,`

`x/2`	`= pi/6, pi-pi/6, 2pi + pi/6, …`
	`= pi/6, (5 pi)/6, (13 pi)/6, …`

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

Trigonometry, 2ADV T2 2007 HSC 4a

Solve `sqrt 2\ sin\ x = 1` for `0 <= x <= 2 pi`. (2 marks)

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`pi/4 , (3 pi)/4`

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`sqrt 2\ sin\ x`	`= 1`	`\ \ \ \ \ \ \ 0 <= x <= 2 pi`
`sin\ x`	`= 1/sqrt 2`

`=>sin\ pi/4 = 1/sqrt 2\ \ text(and sin is positive in)`

`text(1st/2nd quadrants,)`

`:. x`	`= pi/4\ ,\ pi – pi/4`
	`= pi/4\ ,\ (3 pi)/4`

Trigonometry, 2ADV T2 2015 HSC 12a

Find the solutions of `2 sin theta = 1` for `0 <= theta <= 2 pi`. (2 marks)

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`pi/6, (5 pi)/6`

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`2 sin theta`	`= 1,\ \ \ \ 0 <= theta <= 2 pi`
`sin theta`	`= 1/2`

`=> sin (pi/6) = 1/2 \ \ text(and sin is positive)`

`text(in the 1st/2nd quadrants)`

`:. theta`	`= pi/6, pi-pi/6`
	`= pi/6, (5 pi)/6`

Trigonometry, 2ADV T2 2005 HSC 2a

Solve `cos\ theta =1/sqrt2` for `0 ≤ theta ≤ 2pi`. (2 marks)

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`pi/4, (7pi)/4`

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`cos\ theta = 1/sqrt2,\ \ \ 0 ≤ theta ≤ 2pi`

`text(S)text(ince)\ cos\ pi/4 = 1/sqrt2,\ \ text(and cos)`

`text(is positive in 1st/4th quadrants)`

`theta`	`= pi/4, 2pi-pi/4`
	`= pi/4, (7pi)/4`

Trigonometry, 2ADV T2 2008 HSC 6a

Solve `2 sin^2 (x/3) = 1` for `-pi <= x <= pi`. (3 marks)

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`x = -(3pi)/4, (3pi)/4`

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♦♦ Although exact data not available, markers specifically mentioned this question was poorly answered.

`2 sin^2 (x/3)`	`= 1\ \ text(for)\ \ -pi <= x <= pi`
`sin^2 (x/3)`	`= 1/2`
`sin (x/3)`	`= +- 1/sqrt2`

MARKER’S COMMENT: Many students had problems adjusting their answer to the given domain, especially when dealing with negative angles.

`text(When)\ sin (x/3) = 1/sqrt2`

`x/3`	`= pi/4, (3pi)/4`
`x`	`= (3pi)/4, (9pi)/4`

`text(When)\ sin (x/3) = – 1/sqrt2`

`x/3`	`= – pi/4, -(3pi)/4`
`x`	`= -(3pi)/4, -(9pi)/4`

`:.\ x = -(3pi)/4\ \ text(or)\ \ (3pi)/4\ \ text(for)\ \ -pi <= x <= pi`

Trigonometry, 2ADV T2 2009 HSC 1e

Find the exact value of `theta` such that `2 cos theta = 1`, where `0 <= theta <= pi/2`. (2 marks)

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`theta = pi/3\ text(radians)`

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`2 cos theta`	`= 1`
`cos theta`	`= 1/2`
`:.\ theta`	`= pi/3,\ \ \ \ 0 <= theta <= pi/2`

Trigonometry, 2ADV T2 2011 HSC 2b

Find the exact values of `x` such that `2sin x =-sqrt3`, where `0 <= x <= 2pi`. (2 marks)

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`x = (4pi)/3,\ (5pi)/3`

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MARKER’S COMMENT: Better responses found the reference angle and then identified the correct quadrants, as shown clearly in the worked solution.

`2sinx`	`=- sqrt3\ \ text(where)\ \ 0 <= x <= 2pi`
`sin x`	`= -sqrt3/2`
`sin (pi/3)`	`= sqrt3/2`

`text(S)text(ince)\ sin x\ text(is negative in)\ 3^text(rd) // 4^text(th)\ text(quadrants)`

`x`	`= pi + pi/3,\ 2pi-pi/3`
	`= (4pi)/3,\ (5pi)/3\ \ text(radians)`

Trigonometry, 2ADV T2 2012 HSC 6 MC

What are the solutions of `sqrt3 tanx = -1` for `0<=x<=2 pi`?

`(2 pi)/3\ text(and)\ (4 pi)/3`
`(2 pi)/3\ text(and)\ (5 pi)/3`
`(5 pi)/6\ text(and)\ (7 pi)/6`
`(5 pi)/6\ text(and)\ (11 pi)/6`

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`D`

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`sqrt3 tanx`	`= -1`
`tanx`	`= -1/sqrt3`

`text(When)\ tanx = 1/sqrt3,\ \ x=pi/6`

`text(S)text(ince)\ tanx\ text(is negative in)\ 2^text(nd) // 4^text(th)\ text(quadrant)`

`:. x`	` = pi\-pi/6,\ 2 pi\-pi/6,\ …`
	`= (5 pi)/6,\ (11 pi)/6`

`=> D`

`:.\ text(cosec)\ theta`	`= 1/(sintheta)`
	`= 1/(35/37)`
	`= 37/35`

`sectheta`	`= 1/(costheta)`
	`= 1/(24/25)`
	`= 25/24`