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Trigonometry, EXT1 T2 2025 HSC 11b

Solve  \(\sin 2 \theta-\sin \theta=0\)  for  \(0 \leq \theta \leq \pi\).   (3 marks)

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\(\theta=0, \dfrac{\pi}{3}\ \text{or}\ \pi\)

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\(\sin\,2\theta-\sin\,\theta\) \(=0\)  
\(2\,\sin\,\theta\,\cos\,\theta-\sin\,\theta\) \(=0\)  
\(\sin\,\theta(2\,\cos\,\theta-1)\) \(=0\)  

 
\(\sin\,\theta=0\ \ \Rightarrow\ \ \theta=0, \pi\)

\(2\,\cos\,\theta-1=0\ \ \Rightarrow\ \ \cos\,\theta=\dfrac{1}{2}\ \ \Rightarrow\ \ \theta=\dfrac{\pi}{3}\)

\(\therefore \theta=0, \dfrac{\pi}{3}\ \text{or}\ \pi.\)

Trigonometry, EXT1 T2 EQ-Bank 7

Prove  \(\dfrac{\sin A}{\cos A+\sin A}+\dfrac{\sin A}{\cos A-\sin A}=\tan 2 A\).   (2 marks)

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\(\text{LHS}\) \(=\dfrac{\sin A}{\cos A+\sin A}+\dfrac{\sin A}{\cos A-\sin A}\)
  \(=\dfrac{\sin A(\cos A-\sin A)+\sin A(\cos A+\sin A)}{(\cos A+\sin A)(\cos A-\sin A)}\)
  \(=\dfrac{\sin A \cos A-\sin ^2 A+\sin A \cos A+\sin ^2 A}{\cos ^2 A-\sin ^2 A}\)
  \(=\dfrac{2 \sin A \cos A}{\cos 2 A}\)
  \(=\dfrac{\sin 2 A}{\cos 2 A}\)
  \(=\tan 2 A\)
Show Worked Solution
\(\text{LHS}\) \(=\dfrac{\sin A}{\cos A+\sin A}+\dfrac{\sin A}{\cos A-\sin A}\)
  \(=\dfrac{\sin A(\cos A-\sin A)+\sin A(\cos A+\sin A)}{(\cos A+\sin A)(\cos A-\sin A)}\)
  \(=\dfrac{\sin A \cos A-\sin ^2 A+\sin A \cos A+\sin ^2 A}{\cos ^2 A-\sin ^2 A}\)
  \(=\dfrac{2 \sin A \cos A}{\cos 2 A}\)
  \(=\dfrac{\sin 2 A}{\cos 2 A}\)
  \(=\tan 2 A\)

Trigonometry, EXT1 T2 EQ-Bank 8

Prove that \(\dfrac{\cos \alpha-\cos (\alpha+2 \beta)}{2 \sin \beta}=\sin (\alpha+\beta)\).   (3 marks)

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\(\text{LHS}\) \(=\dfrac{\cos \alpha-\cos (\alpha+2 \beta)}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-[\cos \alpha\, \cos 2 \beta+\sin \alpha\, \sin 2 \beta]}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\left[\cos \alpha\left(\cos ^2 \beta-\sin ^2 \beta\right)+\sin \alpha(2 \sin \beta\, \cos \beta)\right]}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\cos \alpha\, \cos ^2 \beta+\cos \alpha\, \sin ^2 \beta+2 \sin \alpha\, \sin \beta\, \cos \beta}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\cos \alpha\left(1-\sin ^2 \beta\right)+\cos \alpha\, \sin ^2 \beta+2 \sin \alpha\, \sin \beta\, \cos \beta}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\cos \alpha+\cos \alpha\, \sin ^2 \beta+\cos \alpha\, \sin ^2 \beta+2 \sin \alpha\, \sin \beta\, \cos \beta}{2 \sin \beta}\)
  \(=\dfrac{2 \sin \beta(\cos \alpha\, \sin \beta+\sin \alpha\, \cos \beta)}{2 \sin \beta}\)
  \(=\sin (\alpha+\beta)\)
Show Worked Solution
\(\text{LHS}\) \(=\dfrac{\cos \alpha-\cos (\alpha+2 \beta)}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-[\cos \alpha\, \cos 2 \beta+\sin \alpha\, \sin 2 \beta]}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\left[\cos \alpha\left(\cos ^2 \beta-\sin ^2 \beta\right)+\sin \alpha(2 \sin \beta\, \cos \beta)\right]}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\cos \alpha\, \cos ^2 \beta+\cos \alpha\, \sin ^2 \beta+2 \sin \alpha\, \sin \beta\, \cos \beta}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\cos \alpha\left(1-\sin ^2 \beta\right)+\cos \alpha\, \sin ^2 \beta+2 \sin \alpha\, \sin \beta\, \cos \beta}{2 \sin \beta}\)
  \(=\dfrac{\cos \alpha-\cos \alpha+\cos \alpha\, \sin ^2 \beta+\cos \alpha\, \sin ^2 \beta+2 \sin \alpha\, \sin \beta\, \cos \beta}{2 \sin \beta}\)
  \(=\dfrac{2 \sin \beta(\cos \alpha\, \sin \beta+\sin \alpha\, \cos \beta)}{2 \sin \beta}\)
  \(=\sin (\alpha+\beta)\)

Trigonometry, EXT1 T2 2024 SPEC2 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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\(A\)

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\(\cos^2(x)+\sin^2(x)\) \(=1\)  
\(\cos^2(x)+a^2\) \(=1\)  
\(\cos(x)\) \(=\sqrt{1-a^2}\ \ \Big(\text{take +ve root since}\ x \in\left(\frac{3 \pi}{2}, 2\pi\right),\ \cos(x)>0 \Big)\)  
\(2\cos^2 \left(\dfrac{x}{2}\right)-1\) \(=\sqrt{1-a^2}\)  
\(\cos^2 \left(\dfrac{x}{2}\right)\) \(=\dfrac{\sqrt{1-a^2}+1}{2}\)  
\(\cos\left(\dfrac{x}{2}\right)\) \(=\pm \sqrt{\dfrac{\sqrt{1-a^2}+1}{2}}\)  

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

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

\(\Rightarrow A\)

♦♦♦ Mean mark 27%.

Trigonometry, EXT1 T2 EQ-Bank 8

  1. Show that  \(\dfrac{\sin 2 x}{1+\cos 2 x}=\tan x\).  (1 mark)

  2. Hence find the exact value of \(\tan \dfrac{\pi}{8}\) in simplest form.  (2 marks)

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a.   \(\dfrac{\sin 2 x}{1+\cos 2 x}\) \(=\dfrac{2\sin x \cos x}{2\cos^{2} x}\)  
  \(=\dfrac{\sin x}{\cos x}\)  
  \(=\tan x\)  

b.   \(\sqrt2-1\)

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a.   \(\dfrac{\sin 2 x}{1+\cos 2 x}\) \(=\dfrac{2\sin x\, \cos x}{2\cos^{2} x}\)  
  \(=\dfrac{\sin x}{\cos x}\)  
  \(=\tan x\)  

 

b.    \(\tan \dfrac{\pi}{8}\) \(=\dfrac{\sin \frac{\pi}{4}}{1+\cos \frac{\pi}{4}}\)
    \(= \dfrac{\frac{1}{\sqrt2}}{1+\frac{1}{\sqrt2}} \times \dfrac{\sqrt2}{\sqrt2} \)
    \(= \dfrac{1}{\sqrt2+1} \times \dfrac{\sqrt2-1}{\sqrt2-1}\)
    \(=\sqrt2-1\)

Trigonometry, EXT1 T2 2013 HSC 8 MC

The angle  `theta`  satisfies  `sin theta = 5/13`  and  `pi/2 < theta < pi`.

What is the value of  `sin 2 theta`?

  1. `10/13`
  2. `- 10/13` 
  3. `120/169`
  4. `- 120/169`
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`D`

Show Worked Solution
 
♦ Mean mark 44% 

`sin theta = 5/13\ \ \ (pi/2 < theta < pi)`

`text(S)text(ince)\ \ theta\ \ text(is in the 2nd quadrant,)`

`cos theta = -12/13`

`:.sin 2theta` `= 2 sin theta cos theta`
  `= 2 xx 5/13 xx -12/13`
  `= -120/169`

`=>  D`