Standard +0.3 This is a standard harmonic form question with routine application. Part (a) uses the standard R sin(x-α) conversion formula (R=√(a²+b²), tan α=b/a). Part (b) requires recognizing that T is minimized when the denominator is maximized, which occurs when R sin(x-α)=R. The steps are mechanical and follow textbook procedures with no novel insight required. Slightly easier than average due to its predictable structure.
6. a. Express \(4 \sin x - 5 \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\).
Give the exact value of \(R\), and give the value of \(\alpha\), in degrees, to 2 decimal places.
$$T = \frac { 8400 } { 19 + ( 4 \sin x - 5 \cos x ) ^ { 2 } } , x > 0$$
b. Use your answer to part \(a\) to calculate
i. the minimum value of \(T\).
ii. the smallest value of \(x , x > 0\), at which this minimum value occurs.
Part a: Express \(4\sin x - 5\cos x\) in the form \(R\sin(x - \alpha)\), where \(R > 0\) and \(0 < \alpha < 90°\). Give the exact value of \(R\), and give the value of \(\alpha\), in degrees, to 2 decimal places
ii. the smallest value of x, x > 0, at which this minimum value occurs
Answer
Marks
M1
Uses \(x - \text{their } \alpha = (2n + 1)90°\) to find x. e.g. \(90° \pm 51.34°\)
A1
\(141.34°\)
(3) marks + (4) marks = (7) marks total for Question 6
**Part a:** Express $4\sin x - 5\cos x$ in the form $R\sin(x - \alpha)$, where $R > 0$ and $0 < \alpha < 90°$. Give the exact value of $R$, and give the value of $\alpha$, in degrees, to 2 decimal places
| $R = \sqrt{4^2 + 5^2} = \sqrt{41}$ only | B1 | $R = \sqrt{4^2 + 5^2} = \sqrt{41}$ only |
| --- | --- | --- |
| $\tan \alpha = \frac{5}{4} \Rightarrow \alpha = 51.34°$ | M1 | Proceeds to a value of $\alpha$ from $\tan \alpha = \pm\frac{5}{4}$, $\tan \alpha = \pm\frac{4}{5}$, or $\cos \alpha = \pm\frac{4}{R}$ |
| | A1 | $\alpha = 51.34°$ or 0.8961 radians |
**Part b:** Use your answer to part a to calculate
**i. the minimum value of T**
| $T = \frac{8400}{19 + (\sqrt{41})^2} = \frac{8400}{60} = 140$ | M1 | for an attempt at $\frac{8400}{19 + (R)^2}$ |
| --- | --- | --- |
| | A1 | 140 |
**ii. the smallest value of x, x > 0, at which this minimum value occurs**
| M1 | Uses $x - \text{their } \alpha = (2n + 1)90°$ to find x. e.g. $90° \pm 51.34°$ |
| --- | --- | --- |
| | A1 | $141.34°$ |
**(3) marks + (4) marks = (7) marks total for Question 6**
---
6. a. Express $4 \sin x - 5 \cos x$ in the form $R \sin ( x - \alpha )$, where $R > 0$ and $0 < \alpha < 90 ^ { \circ }$.
Give the exact value of $R$, and give the value of $\alpha$, in degrees, to 2 decimal places.
$$T = \frac { 8400 } { 19 + ( 4 \sin x - 5 \cos x ) ^ { 2 } } , x > 0$$
b. Use your answer to part $a$ to calculate\\
i. the minimum value of $T$.\\
ii. the smallest value of $x , x > 0$, at which this minimum value occurs.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q6 [7]}}