The function $f(x) = ext{sin} x + ext{cos} x - x$ has a zero near $x = 1.2$. Use one application of Newton's method to find a second approximation to the zero. Write your answer correct to three significant figures. Two circles, $C_1$ and $C_2$, intersect at points A and B. Circle $C_1$ passes through the centre O of circle $C_2$. The point P lies on circle $C_2$ so that the line PA is tangent to circle $C_1$ at point A. Let $ZAB = \theta$. Copy or trace the diagram into your writing booklet. (i) Find $\angle AOB$ in terms of $\theta$. Give a reason for your answer. (ii) Explain why $\angle ZAB = 2\theta$. (iii) Deduce that $PA = BA$. Starting from the identity $\text{sin}(\theta + 2\pi) = \text{sin} 2\theta + \text{cos} 2\theta \cdot \text{sin} 2\theta$, and using the double angle formula, prove the identity $\text{sin} 3\theta = 3\text{sin} \theta - 4\text{sin}^3 \theta$. Hence solve the equation $\text{sin} 3\theta = 2\text{sin} \theta$ for $0 \leq \theta < 2\pi$.

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The function $f(x) = ext{sin} x + ext{cos} x - x$ has a zero near $x = 1.2$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2001 - Paper 1

Question 3

The function $f(x) = ext{sin} x + ext{cos} x - x$ has a zero near $x = 1.2$. Use one application of Newton's method to find a second approximation to the zero. Wr... show full transcript

Worked Solution & Example Answer:The function $f(x) = ext{sin} x + ext{cos} x - x$ has a zero near $x = 1.2$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2001 - Paper 1

Step 1

Find $\angle AOB$ in terms of $\theta$. Give a reason for your answer.

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Answer

To find the angle $\angle AOB$ in terms of $\theta$ , we note that triangles formed by the centre O and points A and B are isosceles, as OA and OB are radii of circle $C_1$ . The angles at A and B, namely $\angle OAP$ and $\angle OBP$ , are equal to $\theta$ . The angle at O, i.e., $\angle AOB$ , therefore, can be calculated as follows:

$\angle AOB = 2\theta.$

This is based on the properties of isosceles triangles, where the angles opposite the equal sides are equal.

Step 2

Explain why $\angle ZAB = 2\theta$.

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Answer

Using the properties of angles formed by tangents and secants, when the tangent PA meets the chord AB, we know that:

$\angle ZAB = \angle AOB$

Thus, substituting the earlier result, we have:

$\angle ZAB = 2\theta.$

The relationship holds because the angle formed at the tangent point to the chord equals the angle subtended at the circumference.

Step 3

Deduce that $PA = BA$.

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Answer

From the above, since $\angle ZAB = 2\theta$ and we also have established that $O$ lies on $C_2$ , we can use properties of tangents. Since line PA is tangent to circle $C_1$ , at A, we can apply the tangent-radius theorem, which states that the tangent at any point is perpendicular to the radius at that point. Thus, with the angles being equal:

$\Rightarrow PA = BA$

indicates that triangles $PAB$ and $OAB$ are congruent, thereby confirming that the lengths PA and BA are equal.

Step 4

Starting from the identity $\text{sin}(\theta + 2\pi) = \text{sin} 2\theta + \text{cos} 2\theta \cdot \text{sin} 2\theta$, prove the identity $\text{sin} 3\theta = 3\text{sin} \theta - 4\text{sin}^3 \theta$.

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Answer

To establish the required identity, recall:

$\text{sin}(3\theta) = \text{sin}(\theta + 2\theta)$

Using the sum of angles formula:

$= \text{sin} \theta \cdot \text{cos}(2\theta) + \text{cos} \theta \cdot \text{sin}(2\theta)$

Now substituting:

$\text{sin}(2\theta) = 2\text{sin} \theta \cdot \text{cos} \theta$ and $\text{cos}(2\theta) = 1 - 2\text{sin}^2 \theta$ ,

we can rewrite the original expression as:

$\text{sin}(3\theta) = \text{sin} \theta (1 - 2\text{sin}^2 \theta) + \text{cos} \theta (2\text{sin} \theta \cdot \text{cos} \theta)$

After simplification, this leads to:

$= 3 \text{sin} \theta - 4\text{sin}^3 \theta,$

which confirms the identity.

Step 5

Hence solve the equation $\text{sin} 3\theta = 2\text{sin} \theta$ for $0 \leq \theta < 2\pi$.

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Answer

Starting from the identity proved earlier:

$3\text{sin} \theta - 4\text{sin}^3 \theta = 2\text{sin} \theta$

Rearranging gives:

$4\text{sin}^3 \theta - \text{sin} \theta = 0$

Factoring out $\text{sin} \theta$ yields:

$\text{sin} \theta (4\text{sin}^2 \theta - 1) = 0$

Setting each factor to zero, we find:

For $\text{sin} \theta = 0$ , solutions are: $\theta = 0, \pi, 2\pi$
For $4\text{sin}^2 \theta - 1 = 0$ : $\text{sin}^2 \theta = \frac{1}{4} \Rightarrow \text{sin} \theta = \frac{1}{2} \text{ or } -\frac{1}{2}$ Leading to: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$

Thus the complete set of solutions is:

$\theta = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi, 2\pi$

Summarizing, the solutions for $0 \leq \theta < 2\pi$ are:

$\theta = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$

The function $f(x) = ext{sin} x + ext{cos} x - x$ has a zero near $x = 1.2$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2001 - Paper 1

Worked Solution & Example Answer:The function $f(x) = ext{sin} x + ext{cos} x - x$ has a zero near $x = 1.2$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2001 - Paper 1

Find $\angle AOB$ in terms of $\theta$. Give a reason for your answer.

Explain why $\angle ZAB = 2\theta$.

Deduce that $PA = BA$.

Starting from the identity $\text{sin}(\theta + 2\pi) = \text{sin} 2\theta + \text{cos} 2\theta \cdot \text{sin} 2\theta$, prove the identity $\text{sin} 3\theta = 3\text{sin} \theta - 4\text{sin}^3 \theta$.

Hence solve the equation $\text{sin} 3\theta = 2\text{sin} \theta$ for $0 \leq \theta < 2\pi$.

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