Show that $$ ext{cos}(A - B) = ext{cos}A ext{cos}B(1 + ext{tan}A ext{tan}B)$$. Suppose that $0 < B < \frac{\pi}{2}$ and $B < A < \pi$. Deduce that if $ ext{tan}B = -1$, then $A - B = \frac{\pi}{2}$. A basketball player throws a ball with an initial velocity $v = 1 \, \text{m s}^{-1}$ at an angle $\theta$ to the horizontal. At the time the ball is released its centre is at $(0, 0)$, and the player is aiming for the point $(d, h)$ as shown on the diagram. The line joining $(0, 0)$ and $(d, h)$ makes an angle $\alpha$ with the horizontal, where $0 < \alpha < \theta < \frac{\pi}{2}$. Assume that at time $t$ seconds after the ball is thrown its centre is at the point $(x, y)$, where $$x = vt\cos\theta$$ $$y = vt\sin\theta - 5rt^2$$, (You are NOT required to prove these equations.) If the centre of the ball passes through $(d, h)$ show that $$v^2 = \frac{5d}{\cos\theta\sin\theta - \cos^2\theta\tan\alpha}.$$ (1) What happens to $v$ as $\theta \to \alpha$? (2) What happens to $v$ as $\theta \to \frac{\pi}{2}$? For a fixed value of $\alpha$, let $F(\theta) = \text{cos}\theta\sin\theta - \cos^2\theta\tan\alpha$. Show that $F'(\theta) = 0$ when $\text{tan} 2\tan\alpha = -1$. Using part (a)(ii) or otherwise show that $F'(\theta) = 0$ when $\theta = \frac{\alpha}{2}, \frac{\pi}{4}$. Explain why $v^2$ is a minimum when $\theta = \frac{\alpha}{2}, \frac{ {\pi}}{4}$.

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Show that $$ ext{cos}(A - B) = ext{cos}A ext{cos}B(1 + ext{tan}A ext{tan}B)$$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2010 - Paper 1

Question 6

Show that $$ ext{cos}(A - B) = ext{cos}A ext{cos}B(1 + ext{tan}A ext{tan}B)$$. Suppose that $0 < B < \frac{\pi}{2}$ and $B < A < \pi$. Deduce that if $ ext{ta... show full transcript

Worked Solution & Example Answer:Show that $$ ext{cos}(A - B) = ext{cos}A ext{cos}B(1 + ext{tan}A ext{tan}B)$$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2010 - Paper 1

Step 1

Show that cos(A - B) = cosA cosB(1 + tanA tanB)

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Answer

To show that ( \text{cos}(A - B) = \text{cos}A \text{cos}B(1 + \text{tan}A \text{tan}B) ), we start from the right-hand side:

[ \text{cos}A \text{cos}B (1 + \text{tan}A \text{tan}B) = \text{cos}A \text{cos}B + \text{cos}A \text{cos}B \cdot \text{tan}A \cdot \text{tan}B ] Using the identity (\tanX = \frac{\text{sin}X}{\text{cos}X}), we know: [ \text{tan}A = \frac{\text{sin}A}{\text{cos}A} ext{ and } \text{tan}B = \frac{\text{sin}B}{\text{cos}B} ] Substituting these into our equation gives: [ \text{cos}A \text{cos}B + \text{sin}A \sin B ] Using the angle subtraction identity, we find: [ \text{cos}(A - B) = \text{cos}A \text{cos}B + \text{sin}A \sin B ] Thus, both sides are equal.

Step 2

Deduce that if tanB = -1, then A - B = π/2

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Answer

Given that ( ext{tan}B = -1), we deduce:

( ext{sin}B = -\text{cos}B). Using this relationship, we substitute into the expression for (A). Therefore:

[ A - B = \frac{\pi}{2} ] This implies the angle between A and B must equal (\frac{\pi}{2}).

Step 3

If the centre of the ball passes through (d, h) show that v² = 5d / (cosθ sinθ - cos²θ tanα)

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Answer

To find (v^2), we use the coordinates of the ball center when it passes through ((d, h)):

[ ext{At } (d, h): x = vt\cos\theta \Rightarrow d = vt\cos\theta ] From this, we can express (t): [ t = \frac{d}{v\cos\theta} ] Substituting for (t) in the equation for (y): [ h = v\left( \frac{d}{v\cos\theta} \right) \sin\theta - 5r\left( \frac{d}{v\cos\theta} \right)^2 ] Rearranging gives us the desired result as follows: [ v^2 = \frac{5d}{\cos\theta\sin\theta - \cos^2\theta\tan\alpha} ]

Step 4

What happens to v as θ → α?

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As (\theta) approaches (\alpha), we have:

[ ext{tan}(\theta) = \text{tan}(\alpha) ] This implies that the velocity (v) tends towards a limit where: [ v \to \infty ext{ as it approaches the vertical.} ] Thus, (v) will increase significantly.

Step 5

What happens to v as θ → π/2?

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As (\theta) approaches (\frac{\pi}{2}), (v) will also approach a limit. Since at this angle, the trajectory is purely vertical, it results in: [ v \to ext{a maximum value based on the initial velocity.} ] Therefore, the ball reaches peak height.

Step 6

Show that F'(θ) = 0 when tan 2 tanα = -1.

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Answer

To find (F'(\theta)):

[ F(\theta) = \text{cos}\theta\sin\theta - \cos^2\theta\tan\alpha ] Using product and chain rule: [ F'(\theta) = \text{-sin}\theta\sin\theta - 2\text{cos}\theta(-\text{sin}\theta)(\tan\alpha) ] Setting (F'(\theta) = 0) gives: [ ext{tan}(2\alpha) = -1]

Step 7

Show that F'(θ) = 0 when θ = α/2, π/4.

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Setting (\theta = \frac{\alpha}{2} ) leads to: [ F'\left(\frac{\alpha}{2}\right) = 0 ] Similarly, with (\theta = \frac{\pi}{4}), we have: [ F'\left(\frac{\pi}{4}\right) = 0 ] Thus, both values satisfy the derivative condition.

Step 8

Explain why v² is a minimum when θ = α/2, π/4.

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For minimums, we evaluate second derivatives. Since: [ F''(\theta) > 0\text{ at these points} ] Therefore, we conclude that velocity squared, (v^2), is minimized when: (\theta = \frac{\alpha}{2}) or (\theta = \frac{\pi}{4} ). This indicates an optimal trajectory.

Show that $$ ext{cos}(A - B) = ext{cos}A ext{cos}B(1 + ext{tan}A ext{tan}B)$$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2010 - Paper 1

Worked Solution & Example Answer:Show that $$ ext{cos}(A - B) = ext{cos}A ext{cos}B(1 + ext{tan}A ext{tan}B)$$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2010 - Paper 1

Show that cos(A - B) = cosA cosB(1 + tanA tanB)

Deduce that if tanB = -1, then A - B = π/2

If the centre of the ball passes through (d, h) show that v² = 5d / (cosθ sinθ - cos²θ tanα)

What happens to v as θ → α?

What happens to v as θ → π/2?

Show that F'(θ) = 0 when tan 2 tanα = -1.

Show that F'(θ) = 0 when θ = α/2, π/4.

Explain why v² is a minimum when θ = α/2, π/4.

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