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Here's the solution to question 3:

3) (i) Show that sin⁡3θ+sin⁡θcos⁡3θ+cos⁡θ=tan⁡2θ\frac{\sin 3\theta + \sin \theta}{\cos 3\theta + \cos \theta} = \tan 2\theta

Step 1: Apply the sum-to-product trigonometric identities.
The identities are:
sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)
cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)

Step 2: Apply the identities to the numerator and denominator.
For the numerator, let A=3θA = 3\theta and B=θB = \theta:

sin⁡3θ+sin⁡θ=2sin⁡(3θ+θ2)cos⁡(3θ−θ2)\sin 3\theta + \sin \theta = 2 \sin\left(\frac{3\theta+\theta}{2}\right) \cos\left(\frac{3\theta-\theta}{2}\right) =2sin⁡(4θ2)cos⁡(2θ2)= 2 \sin\left(\frac{4\theta}{2}\right) \cos\left(\frac{2\theta}{2}\right) =2sin⁡(2θ)cos⁡(θ)= 2 \sin(2\theta) \cos(\theta)

For the denominator, let A=3θA = 3\theta and B=θB = \theta:

cos⁡3θ+cos⁡θ=2cos⁡(3θ+θ2)cos⁡(3θ−θ2)\cos 3\theta + \cos \theta = 2 \cos\left(\frac{3\theta+\theta}{2}\right) \cos\left(\frac{3\theta-\theta}{2}\right) =2cos⁡(4θ2)cos⁡(2θ2)= 2 \cos\left(\frac{4\theta}{2}\right) \cos\left(\frac{2\theta}{2}\right) =2cos⁡(2θ)cos⁡(θ)= 2 \cos(2\theta) \cos(\theta)

Step 3: Substitute the simplified expressions back into the original fraction.

sin⁡3θ+sin⁡θcos⁡3θ+cos⁡θ=2sin⁡(2θ)cos⁡(θ)2cos⁡(2θ)cos⁡(θ)\frac{\sin 3\theta + \sin \theta}{\cos 3\theta + \cos \theta} = \frac{2 \sin(2\theta) \cos(\theta)}{2 \cos(2\theta) \cos(\theta)}

Step 4: Cancel common terms and simplify.
Assuming cos⁡(θ)≠0\cos(\theta) \neq 0, we can cancel 2cos⁡(θ)2 \cos(\theta) from the numerator and denominator:

=sin⁡(2θ)cos⁡(2θ)= \frac{\sin(2\theta)}{\cos(2\theta)} =tan⁡(2θ)= \tan(2\theta)

Thus, it is shown that sin⁡3θ+sin⁡θcos⁡3θ+cos⁡θ=tan⁡2θ\boxed{\frac{\sin 3\theta + \sin \theta}{\cos 3\theta + \cos \theta} = \tan 2\theta}.

3) (ii) Given that f(x)=cos⁡x+3sin⁡xf(x) = \cos x + \sqrt{3} \sin x

Part a· 4 steps

Express f(x)f(x) in the form Rcos⁡(x−λ)R \cos(x - \lambda), where R>0R > 0 and 0<λ<π20 < \lambda < \frac{\pi}{2}.

  1. Use the R-formula for acos⁡x+bsin⁡x=Rcos⁡(x−λ)a \cos x + b \sin x = R \cos(x - \lambda).

    Here, a=1a=1 and b=3b=\sqrt{3}.
    The formula states R=a2+b2R = \sqrt{a^2 + b^2} and tan⁡λ=ba\tan \lambda = \frac{b}{a}.

  2. Calculate RR.

    R=12+(3)2R = \sqrt{1^2 + (\sqrt{3})^2} R=1+3R = \sqrt{1 + 3} R=4R = \sqrt{4} R=2R = 2

  3. Calculate λ\lambda.

    tan⁡λ=31\tan \lambda = \frac{\sqrt{3}}{1} tan⁡λ=3\tan \lambda = \sqrt{3}

    Since a=1a=1 and b=3b=\sqrt{3} are both positive, λ\lambda is in the first quadrant.

    λ=arctan⁡(3)\lambda = \arctan(\sqrt{3}) λ=π3\lambda = \frac{\pi}{3}

    This value satisfies 0<λ<π20 < \lambda < \frac{\pi}{2}.

  4. Write f(x)f(x) in the required form.

    f(x)=2cos⁡(x−π3)f(x) = \boxed{2 \cos\left(x - \frac{\pi}{3}\right)}

Part b· 4 steps

Find the minimum value of 11+∣f(x)∣\frac{1}{1+|f(x)|}.

  1. Determine the range of f(x)f(x).

    From part (a), f(x)=2cos⁡(x−π3)f(x) = 2 \cos\left(x - \frac{\pi}{3}\right).
    The range of cos⁡(x−π3)\cos\left(x - \frac{\pi}{3}\right) is [−1,1][-1, 1].
    Therefore, the range of f(x)f(x) is 2×[−1,1]=[−2,2]2 \times [-1, 1] = [-2, 2].

  2. Determine the maximum value of $|f(x)|$.

    The maximum value of ∣f(x)∣|f(x)| occurs when f(x)=2f(x) = 2 or f(x)=−2f(x) = -2.
    So, max⁡(∣f(x)∣)=∣2∣=2\max(|f(x)|) = |2| = 2.

  3. Find the maximum value of the denominator 1+∣f(x)∣1+|f(x)|.

    To minimize the fraction 11+∣f(x)∣\frac{1}{1+|f(x)|}, we need to maximize its denominator.
    Maximum value of 1+∣f(x)∣=1+max⁡(∣f(x)∣)=1+2=31+|f(x)| = 1 + \max(|f(x)|) = 1 + 2 = 3.

  4. Calculate the minimum value of the expression.

    The minimum value of 11+∣f(x)∣\frac{1}{1+|f(x)|} is 1maximum value of (1+∣f(x)∣)\frac{1}{\text{maximum value of } (1+|f(x)|)}.
    Minimum value =13= \frac{1}{3}.

    The minimum value of 11+∣f(x)∣\frac{1}{1+|f(x)|} is 13\boxed{\frac{1}{3}}.

Part c· 5 steps

Find the general solution of f(x)=3f(x) = \sqrt{3}.

  1. Substitute the R-form of f(x)f(x) into the equation.

    2cos⁡(x−π3)=32 \cos\left(x - \frac{\pi}{3}\right) = \sqrt{3}

  2. Isolate the cosine term.

    cos⁡(x−π3)=32\cos\left(x - \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}

  3. Find the principal value for the angle.

    The principal value for which cos⁡θ=32\cos \theta = \frac{\sqrt{3}}{2} is θ=π6\theta = \frac{\pi}{6}.

  4. Write the general solution for the cosine equation.

    The general solution for cos⁡α=cos⁡β\cos \alpha = \cos \beta is α=2nπ±β\alpha = 2n\pi \pm \beta, where nn is an integer.
    So, x−π3=2nπ±π6x - \frac{\pi}{3} = 2n\pi \pm \frac{\pi}{6}.

  5. Solve for xx in both cases.

    Case 1: x−π3=2nπ+π6x - \frac{\pi}{3} = 2n\pi + \frac{\pi}{6}

    x=2nπ+π6+π3x = 2n\pi + \frac{\pi}{6} + \frac{\pi}{3} x=2nπ+π6+2π6x = 2n\pi + \frac{\pi}{6} + \frac{2\pi}{6} x=2nπ+3π6x = 2n\pi + \frac{3\pi}{6} x=2nπ+π2x = 2n\pi + \frac{\pi}{2}

    Case 2: x−π3=2nπ−π6x - \frac{\pi}{3} = 2n\pi - \frac{\pi}{6}

    x=2nπ−π6+π3x = 2n\pi - \frac{\pi}{6} + \frac{\pi}{3} x=2nπ−π6+2π6x = 2n\pi - \frac{\pi}{6} + \frac{2\pi}{6} x=2nπ+π6x = 2n\pi + \frac{\pi}{6}

    The general solution is x=2nπ+π2 or x=2nπ+π6, where n∈Z\boxed{x = 2n\pi + \frac{\pi}{2} \text{ or } x = 2n\pi + \frac{\pi}{6}, \text{ where } n \in \mathbb{Z}}.

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