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Trigonometry, SPEC2 2024 VCAA 4 MC

Given that  \(\sin (x)=a\),  where  \(x \in\left(\dfrac{3 \pi}{2}, 2 \pi\right)\),  then  \(\cos \left(\dfrac{x}{2}\right)\)  is equal to

  1. \(-\dfrac{\sqrt{1+\sqrt{1-a^2}}}{\sqrt{2}}\)
  2. \(\dfrac{\sqrt{1-\sqrt{a^2-1}}}{\sqrt{2}}\)
  3. \(\dfrac{\sqrt{1+\sqrt{1-a^2}}}{\sqrt{2}}\)
  4. \(-\dfrac{\sqrt{\sqrt{1-a^2}-1}}{\sqrt{2}}\)
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\(A\)

Show Worked Solution
\(\cos^2(x)+\sin^2(x)\) \(=1\)  
\(\cos^2(x)+a^2\) \(=1\)  
\(\cos(x)\) \(=\sqrt{1-a^2}\)  
\(2\cos^2 \left(\dfrac{x}{2}\right)-1\) \(=\sqrt{1-a^2}\)  
\(\cos^2 \left(\dfrac{x}{2}\right)\) \(=\dfrac{\sqrt{1-a^2}+1}{2}\)  
  \(=\sqrt{\dfrac{\sqrt{1-a^2}+1}{2}}\)  

 
\(x \in\left(\dfrac{3 \pi}{2}, 2 \pi\right)\ \ \Rightarrow \ \ \dfrac{x}{2} \in\left(\dfrac{3 \pi}{4}, \pi\right) \)

\(\cos \left( \dfrac{x}{2} \right) \lt 0\ \ \text{(take negative root)}\)

\(\Rightarrow A\)

♦♦♦ Mean mark 27%.

Algebra, SPEC2 2022 VCAA 2 MC

The expression `1-\frac{4\sin^2(x)}{\tan^2(x)+1}` simplifies to

  1. `sin(x) \cos(x)`
  2. `1-2\cos^2(2x)`
  3. `2\sin(2x)`
  4. `2\sin^2(2x)`
  5. `\cos^2(2x)`
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`E`

Show Worked Solution
`1-\frac{4 \sin ^2 x}{\tan ^2 x+1}` `= 1-\frac{4 \sin ^2 x}{\sec ^2 x}`  
  `= 1-(2 \sin x\ \cos x)^2`  
  `= 1-\sin ^2 (2 x)`  
  `= \cos ^2 (2 x)`  

 
`=>E`

Trigonometry, SPEC1 2016 VCAA 9

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

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`-1/5`

Show Worked Solution

`cos(x+y)= cos(x) cos(y) – sin(x) sin(y)`
 

`cos (x) cos (y) + sin (x) sin (y) = 3/5\ \ …\ (1)`

`(sin(x) sin(y))/(cos(x) cos(y))` `=2`  
`sin(x) sin(y)` `= 2 cos (x) cos(y)\ \ …\ (2)`  

  
`text{Substitute (2) into (1):}`

`cos(x) cos(y) + 2 cos(x) cos(y)` `= 3/5`
`3 cos(x) cos(y)` `= 3/5`
`cos(x) cos(y)` `= 1/5`

 
`text(Substitute)\ \ cos(x)cos(y)=1/5\ \ text{into (1):}`

`1/5 + sin(x) sin(y)` `= 3/5`
`sin(x) sin(y)` `= 2/5`

 

`:. cos(x + y)` `= cos(x) cos(y) – sin(x) sin(y)`
  `= 1/5 – 2/5`
  `= -1/5`

Trigonometry, SPEC2-NHT 2017 VCAA 4 MC

If  `sin(theta + phi) = a`  and  `sin(theta - phi) = b`, then  `sin(theta) cos(phi)`  is equal to

  1. `ab`
  2. `sqrt(a^2 + b^2)`
  3. `sqrt (ab)`
  4. `sqrt(a^2 - b^2)`
  5. `(a + b)/2`
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`E`

Show Worked Solution

`sin(theta + phi) = a`

`sin theta cos phi + sin phi cos theta` `= a\ \ …\ (1)`
`sin(theta – phi) = b`  
`sin theta cos phi – sin phi cos theta` `= b\ \ …\ (2)`

 
`(1) + (2):`

`2 sin theta cos phi` `= a + b`
`:. sin theta cos phi` `= (a + b)/2`

 
`=>   E`

Trigonometry, SPEC1 2018 VCAA 7

Given that  \(\cot (2 x)+\dfrac{1}{2}\, \tan (x)=a \cot (x)\), use a suitable double angle formula to find the value of \(a , a\) in RR. (3 marks)

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\(a=\dfrac{1}{2}\)

Show Worked Solution

\(\dfrac{1}{\tan (2 x)}+\dfrac{\tan (x)}{2}=\dfrac{a}{\tan (x)}, \quad \tan (x) \neq 0\)

\(\dfrac{1-\tan ^2(x)}{2 \tan (x)}+\dfrac{\tan (x)}{2}=\dfrac{a}{\tan (x)}\)

 
\(\text{If \(\tan(x) \neq 0\):}\)

\(\dfrac{1-\tan ^2(x)+\tan ^2(x)}{2 \tan (x)}\) \(=\dfrac{a}{\tan (x)}\)
\(1\) \(=2 a\)
\(\therefore a\) \(=\dfrac{1}{2}\)