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Trigonometry, EXT1 T3 EQ-Bank 6

Given  \(t=\tan \dfrac{x}{2}\),  prove that  \(\dfrac{\sec x+\tan x}{\sec x-\tan x}=\left(\dfrac{1+t}{1-t}\right)^2\)   (2 marks)

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  \(\operatorname{LHS}\) \( =\dfrac{\sec x+\tan x}{\sec x-\tan x}\)
    \( =\dfrac{\frac{1+t^2}{1-t^2}+\frac{2 t}{1-t^2}}{\frac{1+t^2}{1-t^2}-\frac{2 t}{1-t^2}} \times \dfrac{1-t^2}{1-t^2}\)
    \( =\dfrac{1+t^2+2 t}{1+t^2-2 t}\)
    \(=\dfrac{(1+t)^2}{(1-t)^2 }\)
    \( =\left(\dfrac{1-t}{1-t}\right)^2\)
Show Worked Solution
  \(\operatorname{LHS}\) \( =\dfrac{\sec x+\tan x}{\sec x-\tan x}\)
    \( =\dfrac{\frac{1+t^2}{1-t^2}+\frac{2 t}{1-t^2}}{\frac{1+t^2}{1-t^2}-\frac{2 t}{1-t^2}} \times \dfrac{1-t^2}{1-t^2}\)
    \( =\dfrac{1+t^2+2 t}{1+t^2-2 t}\)
    \(=\dfrac{(1+t)^2}{(1-t)^2 }\)
    \( =\left(\dfrac{1-t}{1-t}\right)^2\)

Trigonometry, EXT1 T3 EQ-Bank 5

By making the substitution  \(t=\tan \dfrac{\theta}{2}\), or otherwise, show that  \(\dfrac{1+\sin x-\cos x}{1+\sin x+\cos x}=\tan \dfrac{x}{2}\).   (3 marks)

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  \(\operatorname{LHS} \) \( =\dfrac{1+\sin x-\cos x}{1+\sin x+\cos x}\)
    \( =\dfrac{1+\dfrac{2 t}{1+t^2}-\dfrac{1-t^2}{1+t^2}}{1+\dfrac{2 t}{1+t^2}+\dfrac{1-t^2}{1+t^2}} \times \dfrac{1+t^2}{1+t^2}\)
    \( =\dfrac{1+t^2+2 t-1+t^2}{1+t^2+2 t+1-t^2}\)
    \( =\dfrac{2 t(t+1)}{2(t+1)} \)
    \(=t\)
    \(=\tan \dfrac{x}{2}\)

Show Worked Solution

  \(\operatorname{LHS} \) \( =\dfrac{1+\sin x-\cos x}{1+\sin x+\cos x}\)
    \( =\dfrac{1+\dfrac{2 t}{1+t^2}-\dfrac{1-t^2}{1+t^2}}{1+\dfrac{2 t}{1+t^2}+\dfrac{1-t^2}{1+t^2}} \times \dfrac{1+t^2}{1+t^2}\)
    \( =\dfrac{1+t^2+2 t-1+t^2}{1+t^2+2 t+1-t^2}\)
    \( =\dfrac{2 t(t+1)}{2(t+1)} \)
    \(=t\)
    \(=\tan \dfrac{x}{2}\)

   

Trigonometry, EXT1 T3 2023 HSC 11e

Solve  \(\cos \theta+\sin \theta=1\)  for  \(0 \leq \theta \leq 2 \pi\).  (3 marks)

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\(\theta=0, \dfrac{\pi}{2}, 2\pi \)

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\(\text{Let}\ \ t=\tan \dfrac{\theta}{2} \)

\(\dfrac{1-t^2}{1+t^2} + \dfrac{2t}{1+t^2} \) \(=1\ \ \text{(see reference sheet)}\)  
\(1-t^2+2t\) \(=1+t^2\)  
\(2t^2-2t\) \(=0\)  
\(2t(t-1) \) \(=0\)  
\(t\) \(= 0 \ \text{or} \ 1\)  

 
\(\text{When}\ \ \tan\dfrac{\theta}{2} = 0: \)

\(\dfrac{\theta}{2} = 0, \pi\ \ \Rightarrow \ \theta=0\ \ \text{or}\ \ 2\pi \)

\(\text{When}\ \ \tan\dfrac{\theta}{2} = 1: \)

\(\dfrac{\theta}{2} = \dfrac{\pi}{4}\ \Rightarrow \ \theta=\dfrac{\pi}{2} \)
 

\(\text{Test}\ \ \theta=\pi: \)

\(\cos\ \pi+\sin\ \pi = -1+0=-1\ \ \text{(not a solution)} \)

\(\therefore \theta=0, \dfrac{\pi}{2}, 2\pi \)

Trigonometry, EXT1 T3 2007 HSC 2a

By using the substitution  `t = tan\ theta/2`, or otherwise, show that  `(1 − cos\ theta)/(sin\ theta) = tan\ theta/2`.  (2 marks)

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`text{Proof (See Worked Solutions)}`

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COMMENT: The values of `sin theta` and `cos theta` can be committed to memory or quickly derived using double angle formulae (as shown in the worked solution).

EXT1 2007 2a

`=>tan\ theta/2 = t`

`=>sin theta` `= 2 · t/(sqrt(1 + t^2)) · 1/(sqrt(1 + t^2)) = (2t)/(1 + t^2)`
`=>cos theta` `= 1/(1 + t^2) − t^2/(1 + t^2) = (1 − t^2)/(1 + t^2)`

 

`text(Show)\ \ (1 − cos theta)/(sin theta) = tan\ theta/2 :`

`(1 − cos\ theta)/(sin theta)` `= (1 − ((1 − t^2)/(1 +  t^2)))/((2t)/(1 + t^2)) xx (1+t^2)/(1+t^2)`
  `= (1 + t^2 − (1 − t^2))/(2t)`
  `= (2t^2)/(2t)`
  `= t`
  `= tan\ theta/2\ \ …text(as required)`

Trigonometry, EXT1 T3 2005 HSC 4b

By making the substitution  `t = tan\ theta/2`, or otherwise, show that

`text(cosec)\ theta + cot theta = cot\ theta/2.`  (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

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`text(Show cosec)\ theta + cot\ theta = cot\ theta/2`
 

`tan\ theta/2=t`

`=>sin theta` `= 2 sin\ theta/2\ cos\ theta/2= (2t)/(t^2 +1)`
`=>cos theta` `= cos^2\ theta/2 – sin^2\ theta/2= (1 – t^2)/(t^2 + 1)`
`=>tan theta` `= (sin theta)/(cos theta)= (2t)/(1 – t^2)`

 

`text(LHS)` `= 1/(sin theta) + 1/(tan theta)`
  `= (t^2 + 1)/(2t) + (1 – t^2)/(2t)`
  `= (t^2 + 1 + 1 – t^2)/(2t)`
  `= 1/t`
  `= 1/(tan\ theta/2)`
  `= cot\ theta/2`
  `=\ text(RHS  …  as required.)`