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Trigonometry, EXT1 T2 EQ-Bank 31

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 EQ-Bank 14

Show that  \(\sin 75^{\circ}=\dfrac{\sqrt{2}+\sqrt{6}}{4}\).   (2 marks)

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\(\sin 75^{\circ}\) \(=\sin \left(30^{\circ}+45^{\circ}\right)\)
  \(=\sin 30^{\circ} \cos 45^{\circ}+\cos 30^{\circ} \sin 45^{\circ}\)
  \(=\dfrac{1}{2} \cdot \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{\sqrt{2}}\)
  \(=\dfrac{1+\sqrt{3}}{2 \sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}\)
  \(=\dfrac{\sqrt{2}+\sqrt{6}}{4}\)
Show Worked Solution
\(\sin 75^{\circ}\) \(=\sin \left(30^{\circ}+45^{\circ}\right)\)
  \(=\sin 30^{\circ} \cos 45^{\circ}+\cos 30^{\circ} \sin 45^{\circ}\)
  \(=\dfrac{1}{2} \cdot \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{\sqrt{2}}\)
  \(=\dfrac{1+\sqrt{3}}{2 \sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}\)
  \(=\dfrac{\sqrt{2}+\sqrt{6}}{4}\)

Trigonometry, EXT1 T2 EQ-Bank 15

Using compound angles, determine the exact value of \(\sin 15^{\circ}\) in its simplest form.   (2 marks)

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\( \sin 15^{\circ}=\dfrac{\sqrt{6}-\sqrt{2}}{4} \)

Show Worked Solution

  \( \sin 15^{\circ}\) \( =\sin (45-30)^{\circ}\)
    \(=\sin 45^{\circ} \, \cos 30^{\circ}-\cos 45^{\circ} \, \sin 30^{\circ} \)
    \(=\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} \)
    \(=\dfrac{\sqrt{3}-1}{2 \sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \)
    \(=\dfrac{\sqrt{6}-\sqrt{2}}{4} \)

Trigonometry, EXT1 T2 EQ-Bank 28

  1. Show that  \(\cos 30^{\circ} \cos 15^{\circ}=\dfrac{1}{2}\left[\cos 15^{\circ}+\dfrac{1}{\sqrt{2}}\right]\).   (2 marks)
  2. Hence, or otherwise, find the exact value of \(\cos 15^{\circ}\).   (2 marks)

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a.   \(\text{Show:}\ \cos 30^{\circ} \cos 15^{\circ}=\dfrac{1}{2}\left[\cos 15^{\circ}+\dfrac{1}{\sqrt{2}}\right]\)

\(\cos(30^{\circ}+15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}-\sin 30^{\circ} \sin 15^{\circ}\ …\ (1)\)

\(\cos(30^{\circ}-15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}+\sin 30^{\circ} \sin 15^{\circ}\ …\ (2)\)

  \(\text{Add (1) + (2):}\)

\(2\cos 30^{\circ} \cos 15^{\circ}\) \(=\cos 45^{\circ}+\cos 15^{\circ}\)  
\(\cos 30^{\circ} \cos 15^{\circ}\) \(=\dfrac{1}{2}\Big[\cos 15^{\circ}+\dfrac{1}{\sqrt2}\Big] \)  

 
b.  \(\cos 15^{\circ}=\dfrac{\sqrt6+\sqrt2}{4} \)

Show Worked Solution

a.   \(\text{Show:}\ \cos 30^{\circ} \cos 15^{\circ}=\dfrac{1}{2}\left[\cos 15^{\circ}+\dfrac{1}{\sqrt{2}}\right]\)

\(\cos(30^{\circ}+15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}-\sin 30^{\circ} \sin 15^{\circ}\ …\ (1)\)

\(\cos(30^{\circ}-15^{\circ})=\cos 30^{\circ} \cos 15^{\circ}+\sin 30^{\circ} \sin 15^{\circ}\ …\ (2)\)

  \(\text{Add (1) + (2):}\)

\(2\cos 30^{\circ} \cos 15^{\circ}\) \(=\cos 45^{\circ}+\cos 15^{\circ}\)  
\(\cos 30^{\circ} \cos 15^{\circ}\) \(=\dfrac{1}{2}\Big[\cos 15^{\circ}+\dfrac{1}{\sqrt2}\Big] \)  

 

b.    \(2\cos 30^{\circ} \cos 15^{\circ}\) \(=\cos 15^{\circ}+\dfrac{1}{\sqrt2}\)
  \(\cos 15^{\circ}(2\cos 30^{\circ}-1)\) \(=\dfrac{1}{\sqrt2}\)
  \(\cos 15^{\circ}(\sqrt3-1)\) \(=\dfrac{1}{\sqrt2}\)
  \(\cos 15^{\circ}\) \(=\dfrac{1}{\sqrt2(\sqrt3-1)}\)
    \(=\dfrac{1}{\sqrt6-\sqrt2} \times \dfrac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}\)
    \(=\dfrac{\sqrt6+\sqrt2}{4} \)

Trigonometry, EXT1 T2 2021 HSC 13d

  1. The numbers `A`, `B` and `C` are related by the equations  `A = B-d`  and  `C = B + d`,  where `d` is a constant.
  2. Show that  `(sin A + sin C)/(cos A + cos C) = tan B`.   (2 marks)

  3. Hence, or otherwise, solve  `(sin\ (5theta)/7 + sin\ (6theta)/7)/(cos\ (5theta)/7 + cos\ (6theta)/7) = sqrt3`  for  `0 <= theta <= 2pi`.   (2 marks)

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  1. `text(See Worked Solution)`
  2. `(14pi)/33, (56pi)/33`
Show Worked Solution
i.    `(sin A + sin C)/(cos A + cos C)` `= (sin (B-d) + sin (B + d))/(cos (B-d) + cos (B + d))`
    `= (2sin B cos d)/(2cos B cosd)`
    `= tan B=\ text(RHS)`

 

ii.   `text(Let)\ \ A = (5theta)/7,\ \ C = (6theta)/7`

♦ Mean mark 50%.

`B= (A + C)/2= 1/2((5theta)/7 + (6theta)/7)= (11theta)/14`

`tan\ ((11theta)/14)` `= sqrt3`
`(11theta)/14` `= pi/3, (4pi)/3`
`:.theta` `= (14pi)/33, (56pi)/33`

Trigonometry, EXT1 T2 EQ-Bank 3 MC

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

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

Show Worked Solution

\(\text{Solution 1}\)

\(\sin \theta \cos \phi=\dfrac{1}{2} [(\sin (\theta+\phi)+\sin (\theta-\phi)]=\dfrac{1}{2}(a+b)\)

\(\Rightarrow D\)
 

\(\text{Solution 2}\)

\(\sin (\theta+\phi)=a\)

\(\sin \theta \cos \phi+\sin \phi \cos \theta\) \(=a \ldots(1)\)
\(\sin (\theta-\phi)=b\)  
\(\sin \theta \cos \phi-\sin \phi \cos \theta\) \(=b \ldots(2)\)

 
\((1)+(2):\)

\(2 \sin \theta \cos \phi\) \(=a+b\)
\(\therefore \sin \theta \cos \phi\) \(=\dfrac{a+b}{2}\)

 

Calculus, EXT1 C2 2019 HSC 14c

The diagram shows the two curves  `y = sin x`  and  `y = sin(x-alpha) + k`, where  `0 < alpha < pi`  and  `k > 0`. The two curves have a common tangent at `x_0` where  `0 < x_0 < pi/2`.
 

  1. Explain why   `cos x_0 = cos (x_0-alpha)`.   (1 mark)

  2. Show that  `sin x_0 = -sin(x_0-alpha)`.   (2 marks)

  3. Hence, or otherwise, find `k` in terms of `alpha`.   (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `k = 2 sin\ alpha/2`
Show Worked Solution
i.    `y_1` `= sin x`
  `(dy_1)/(dx)` `= cos x`
  `y_2` `= sin(x-alpha) + k`
  `(dy_2)/(dx)` `= cos (x-alpha)`

 
`text(At)\ \ x = x_0,\ \ text(tangent is common)`

♦ Mean mark part (i) 47%.

`:. cos x_0 = cos(x_0-alpha)`
 

ii.   `x_0\ text{is in 1st quadrant (given).}`

`text{Using part  (i):}`

`cos\ x_0 = cos(x_0-alpha) >0`

♦♦♦ Mean mark part (ii) 19%.

`=> x_0-alpha\ text(is in 4th quadrant)\ (0 < alpha < pi)`

`text(S)text(ince sin is positive in 1st quadrant and)`

`text(negative in 4th quadrant)`

`=> sin x_0 = -sin(x_0-alpha)`

 

iii.   

`text(When)\ \ x = x_0:`

`y_1` `=sin x_0`  
`y_2` `=sin(x_0-alpha) + k`  
`sin x_0` `=sin (x_0-alpha) + k`  
`sin x_0` `= -sin x_0 + k`  
`k` `== 2\ sin x_0`  

 

♦♦ Mean mark part (iii) 21%.

`text(S)text(ince)\ \ cos x_0` `= cos(x_0-alpha)`
`x_0` `= -(x_0-alpha)`
`2x_0` `= alpha`
`x_0` `= alpha/2`

 
 `:. k = 2 sin\ alpha/2`

Trigonometry, EXT1 T2 EQ-Bank 22

Find the exact value of `cos((11pi)/12)`.   (2 marks)

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`-((sqrt2 + sqrt6))/4`

Show Worked Solution
`cos((11pi)/12)` `= cos((2pi)/3 + pi/4)`
  `= cos((2pi)/3) · cos(pi/4)-sin((2pi)/3) · sin(pi/4)`
  `= cos(pi-pi/3) · 1/sqrt2-sin (pi-pi/3) · 1/sqrt2`
  `= -1/2 · 1/sqrt2-sqrt3/2 · 1/sqrt2`
  `= -((1 + sqrt3))/(2sqrt2)`
  `= -((sqrt2 + sqrt6))/4`

Trigonometry, EXT1 T2 EQ-Bank 23

Find the exact value of `cos\ pi/8`.   (2 marks)

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`(sqrt(sqrt2 + 2))/2`

Show Worked Solution

`text(Using)\ \ cos2A = 2cos^2A-1:`

`2cos^2\ pi/8-1` `= cos\ pi/4`
`2cos^2\ pi/8` `= 1/sqrt2 + 1`
`cos^2\ pi/8` `= (1 + sqrt2)/(2sqrt2) xx sqrt2/sqrt2= (sqrt2 + 2)/4`
`:. cos\ pi/8` `= sqrt((sqrt2 + 2)/4)= (sqrt(sqrt2 + 2))/2`

Trigonometry, EXT1 T2 EQ-Bank 16

Find `a` and `b` such that

`tan75^@ = a + bsqrt3`   (2 marks)

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`a = 2, b = 1`

Show Worked Solution
`tan75^@` `= tan(45 + 30)^@`
  `= (tan45^@ + tan30^@)/(1-tan45^@tan30^@)`
  `= (1 + 1/sqrt3)/(1-1 · 1/sqrt3) xx sqrt3/sqrt3`
  `= (sqrt3 + 1)/(sqrt3-1) xx (sqrt3 + 1)/(sqrt3 + 1)`
  `= (3 + 2sqrt3 + 1)/((sqrt3)^2-1^2)`
  `= 2 + sqrt3`

 
`:. a = 2, \ b = 1`