SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Trigonometry, EXT1 T2 EQ-Bank 8

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

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

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

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

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

Show Answers Only

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

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 3

  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)

Show Answers Only

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)

Show Answers Only
  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)`
    `= (sin B)/(cos B)`
    `= 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 SM-Bank 9 MC

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

A.   `sqrt(a^2 + b^2)`

B.   `sqrt (ab)`

C.   `sqrt(a^2 - b^2)`

D.   `(a + b)/2`

Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

`sintheta cos phi` `=1/2[(sin(theta+phi)+sin(theta – phi)]`   
  `=1/2(a+b)`  

`=>D`

 

`text(Solution 2)`

`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`

 

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)

Show Answers Only
  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 + 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 7

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

Show Answers Only

`−((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 SM-Bank 2

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

Show Answers Only

`(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 1

Find  `a`  and  `b`  such that

`tan75^@ = a + bsqrt3`   (2 marks)

Show Answers Only

`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`