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Given that θ is measured in radians, prove, from first principles, that the derivative of sin θ is cos θ - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1
Question 10
Given that θ is measured in radians, prove, from first principles, that the derivative of sin θ is cos θ. You may assume the formula for sin(A ± B) and that as h → ... show full transcript
Worked Solution & Example Answer:Given that θ is measured in radians, prove, from first principles, that the derivative of sin θ is cos θ - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1
Step 1
Use of \( \sin(θ + h) - \sin(θ) \)
Answer
To find the derivative of ( \sin θ ), we start with the definition of the derivative: [ \frac{d}{dθ} \sin θ = \lim_{h \to 0} \frac{\sin(θ + h) - \sin(θ)}{h} ] Using the sine addition formula: ( \sin(θ + h) = \sin θ \cos h + \cos θ \sin h ) yields: [ \sin(θ + h) - \sin θ = \sin θ \cos h + \cos θ \sin h - \sin θ = \sin θ (\cos h - 1) + \cos θ \sin h ]
Step 2
Substituting into the limit expression
Answer
Now substituting back into the limit expression, we get: [ \frac{d}{dθ} \sin θ = \lim_{h \to 0} \frac{\sin θ (\cos h - 1) + \cos θ \sin h}{h} ] We can separate this into two limits: [ \frac{d}{dθ} \sin θ = \lim_{h \to 0} \left(\frac{\sin θ (\cos h - 1)}{h} + \frac{\cos θ \sin h}{h}\right) ]
Step 3
Step 4