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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 12 - 2017 - Paper 1

Question 12

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 → 0... 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 12 - 2017 - Paper 1

Step 1

Use of \( \sin(θ + h) - \sin θ \)

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Answer

Start with the difference quotient:

[ \frac{\sin(θ + h) - \sin θ}{h} ]

Using the sine addition formula:

[ \sin(θ + h) = \sin θ \cos h + \cos θ \sin h ]

Substituting this into the difference quotient gives:

[ \frac{\sin θ \cos h + \cos θ \sin h - \sin θ}{h} = \frac{\sin θ (\cos h - 1) + \cos θ \sin h}{h} ]

Step 2

Set limits as \( h \to 0 \)

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Answer

As ( h \to 0 ), we know:

( \frac{\sin h}{h} \to 1 )
( \frac{\cos h - 1}{h} \to 0 )

Therefore:

[ \lim_{h \to 0} \frac{\sin(θ + h) - \sin θ}{h} = \sin θ \cdot 0 + \cos θ \cdot 1 = \cos θ ]

Step 3

Conclude the proof

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Answer

Thus, we have shown that the derivative of ( \sin θ ) is ( \cos θ ), which completes the proof from first principles.

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