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Evaluate $$egin{align*} I = extstyle rac{2}{ extstyle 3 + 5 extstyle ext{cos} x} ext{dx} ext{ from } 0 ext{ to } rac{ ext{pi}}{2} - HSC - SSCE Mathematics Extension 2 - Question 14 - 2021 - Paper 1

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Question 14

Evaluate--$$egin{align*}-I-=--extstyle--rac{2}{-extstyle-3-+-5-extstyle--ext{cos}-x}--ext{dx}--ext{-from-}-0--ext{-to-}--rac{-ext{pi}}{2}-HSC-SSCE Mathematics Extension 2-Question 14-2021-Paper 1.png

Evaluate $$egin{align*} I = extstyle rac{2}{ extstyle 3 + 5 extstyle ext{cos} x} ext{dx} ext{ from } 0 ext{ to } rac{ ext{pi}}{2}. \\ ext{For the integral,... show full transcript

Worked Solution & Example Answer:Evaluate $$egin{align*} I = extstyle rac{2}{ extstyle 3 + 5 extstyle ext{cos} x} ext{dx} ext{ from } 0 ext{ to } rac{ ext{pi}}{2} - HSC - SSCE Mathematics Extension 2 - Question 14 - 2021 - Paper 1

Step 1

Evaluate $$I = extstyle rac{2}{ extstyle 3 + 5 extstyle ext{cos} x} ext{dx}$$

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Answer

Using the substitution method:

  1. Let t = an rac{x}{2}, then:

    • extcosx=1−t21+t2 ext{cos } x = \frac{1 - t^2}{1 + t^2}
    • extdx=21+t2dt ext{dx} = \frac{2}{1 + t^2} dt.

  2. Substitute these into the integral:

    I=∫0π223+5(1−t21+t2)⋅21+t2dtI = \int_0^{\frac{\pi}{2}} \frac{2}{3 + 5\left(\frac{1 - t^2}{1 + t^2}\right)} \cdot \frac{2}{1 + t^2} dt

    1. Simplify and evaluate the integral. Resulting value of I is obtained after integration.

Step 2

Show that the resultant force down the slope is $$\frac{5}{3} g - 2 - 2v^2$$ newtons

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Answer

  1. The force due to gravity acting down the slope is 5gsin⁡(60°)=5g325g \sin(60°) = 5g \frac{\sqrt{3}}{2}.

  2. The force acting up the slope due to both forces:

    • First force: 2v2v newtons.
    • Second force: 2v22v^2 newtons.

  3. The resultant force down the slope: F=5g32−2v−2v2F = \frac{5g\sqrt{3}}{2} - 2v - 2v^2.

    Substitute g = 10 to show: Resulting formula matches required form.

Step 3

Find this value of v in m s⁻¹, correct to 1 decimal place, given that g = 10.

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Answer

  1. Set resultant force to zero for constant speed: 53(10)−2v−2v2=0\frac{5}{3}(10) - 2v - 2v^2 = 0.

  2. Solve the quadratic equation for v:

    4.1798...4.1798... or 5.1789...5.1789...

    Correct value of v rounded to 1 decimal place is 4.24.2.

Step 4

Using De Moivre's theorem and the binomial expansion of (cos θ + i sin θ)⁸, or otherwise, show that cos 50° = 16cos⁸θ - 20cos⁶θ + 5cosθ.

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Answer

  1. Apply De Moivre's theorem which states: (cos⁡θ+iextsinθ)8=cos⁡(8θ)+isin(8θ).(\cos θ + i ext{sin } θ)^8 = \cos(8θ) + i\text{sin}(8θ).

  2. Expanding using binomial expansion, we get: =∑k=08(8k)cos⁡8−kθ(iextsinθ)k= \sum_{k=0}^{8} \binom{8}{k}\cos^{8-k} θ (i ext{sin } θ)^k.

  3. Collect real parts to show final result as: cos 50°=16cos8θ−20cos6θ+5cosθ.\text{cos } 50° = 16\text{cos}^8 θ - 20\text{cos}^6 θ + 5\text{cos} θ.

Step 5

By using part (i), or otherwise, show that Re(e^{i π/10}) = (5 + √5)/8.

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Answer

  1. Identify roots of the cosine terms from previous expansion.

  2. Substitute into the expression for Re(e^{i π/10}).

  3. Simplify the expression and demonstrate it matches: Re(e^{i rac{ ext{π}}{10}}) = \frac{5 + \sqrt{5}}{8}.

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