Standard +0.8 This is a from-first-principles proof requiring careful algebraic manipulation of the difference quotient, application of the sin(A+B) formula, and correct use of two standard limits. While the result is standard, executing the proof rigorously with proper limit handling is more demanding than routine differentiation exercises, placing it moderately above average difficulty.
10. Given that \(\theta\) is measured in radians, prove, from first principles, that the derivative of \(\sin \theta\) is \(\cos \theta\)
You may assume the formula for \(\sin ( A \pm B )\) and that as \(h \rightarrow 0 , \frac { \sin h } { h } \rightarrow 1\) and \(\frac { \cos h - 1 } { h } \rightarrow 0\)
Use of \(\frac{\sin(\theta+h)-\sin\theta}{(\theta+h)-\theta}\)
B1
States/implies gradient of chord is \(\frac{\sin(\theta+h)-\sin\theta}{h}\)
Uses compound angle identity for \(\sin(A+B)\) with \(A=\theta\), \(B=h\) \(\Rightarrow \sin(\theta+h) = \sin\theta\cos h + \cos\theta\sin h\)
M1
Uses compound angle identity
\(\frac{\sin(\theta+h)-\sin\theta}{h} = \frac{\sin\theta\cos h + \cos\theta\sin h - \sin\theta}{h}\)
A1
Obtains correct expression
\(= \frac{\sin h}{h}\cos\theta + \left(\frac{\cos h - 1}{h}\right)\sin\theta\)
M1
Writes expression in terms of \(\frac{\sin h}{h}\) and \(\frac{\cos h - 1}{h}\)
Uses \(h \to 0\), \(\frac{\sin h}{h} \to 1\) and \(\frac{\cos h - 1}{h} \to 0\); hence \(\lim_{h\to 0}\frac{\sin(\theta+h)-\sin\theta}{(\theta+h)-\theta} = \cos\theta\) and gradient of chord \(\to\) gradient of curve \(\Rightarrow \frac{dy}{d\theta} = \cos\theta\)
A1*
Uses correct language to explain \(\frac{dy}{d\theta} = \cos\theta\); must use all given statements and explain chord \(\to\) curve transition
Alternative Method (10alt):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Use of \(\frac{\sin(\theta+h)-\sin\theta}{(\theta+h)-\theta}\)
B1
As above
Sets \(\frac{\sin(\theta+h)-\sin\theta}{(\theta+h)-\theta} = \frac{\sin\left(\theta+\frac{h}{2}+\frac{h}{2}\right)-\sin\left(\theta+\frac{h}{2}-\frac{h}{2}\right)}{h}\) and uses compound angle identities for \(\sin(A+B)\) and \(\sin(A-B)\) with \(A=\theta+\frac{h}{2}\), \(B=\frac{h}{2}\)
10. Given that $\theta$ is measured in radians, prove, from first principles, that the derivative of $\sin \theta$ is $\cos \theta$
You may assume the formula for $\sin ( A \pm B )$ and that as $h \rightarrow 0 , \frac { \sin h } { h } \rightarrow 1$ and $\frac { \cos h - 1 } { h } \rightarrow 0$\\
\hfill \mbox{\textit{Edexcel Paper 1 Q10 [5]}}