A-Level Edexcel Maths Pure Partial Fractions: Given that θ is measured in radi

Given that θ is measured in radians, prove, from first principles, that d dθ (cosθ) = -sinθ You may assume the formula for cos(A ± B) and that as h → 0, sin(h)/h → 1 and cos(h) - 1/h → 0. - Edexcel - A-Level Maths Pure - Question 11 - 2018 - Paper 2

Question 11

Given-that-θ-is-measured-in-radians,-prove,-from-first-principles,-that--d--dθ-(cosθ)-=--sinθ--You-may-assume-the-formula-for-cos(A-±-B)-and-that-as-h-→-0,-sin(h)/h-→-1-and-cos(h)---1/h-→-0.-Edexcel-A-Level Maths Pure-Question 11-2018-Paper 2.png

Given that θ is measured in radians, prove, from first principles, that d dθ (cosθ) = -sinθ You may assume the formula for cos(A ± B) and that as h → 0, sin(h)/h ... show full transcript

Worked Solution & Example Answer:Given that θ is measured in radians, prove, from first principles, that d dθ (cosθ) = -sinθ You may assume the formula for cos(A ± B) and that as h → 0, sin(h)/h → 1 and cos(h) - 1/h → 0. - Edexcel - A-Level Maths Pure - Question 11 - 2018 - Paper 2

Step 1

Using the Definition of the Derivative

96%

114 rated

Answer

To prove the statement, we start with the definition of the derivative:

rac{d}{dθ}( ext{cos}θ) = rac{ ext{lim}_{h o 0}( ext{cos}(θ+h) - ext{cos}θ)}{h}

The formula for the cosine of a sum provides:

ext{cos}(θ + h) = ext{cos}θ ext{cos}h - ext{sin}θ ext{sin}h.

Substituting this into our limit expression gives:

rac{ ext{lim}_{h o 0} ( ext{cos}θ ext{cos}h - ext{sin}θ ext{sin}h - ext{cos}θ)}{h}.

Step 2

Simplifying the Expression

99%

104 rated

Answer

Now rearranging the expression gives:

ext{lim}_{h o 0} rac{( ext{cos}θ ( ext{cos}h - 1) - ext{sin}θ ext{sin}h)}{h}.

We can separate this into two limits:

= ext{cos}θ rac{ ext{lim}_{h o 0} ( ext{cos}h - 1)}{h} - ext{sin}θ rac{ ext{lim}_{h o 0} ext{sin}h}{h}.

Step 3

Using Known Limit Results

96%

101 rated

Answer

As given, we know that:

$rac{ ext{sin}h}{h} o 1$ as $h o 0$ .
$rac{ ext{cos}h - 1}{h} o 0$ as $h o 0$ . Thus:

$rac{d}{dθ}( ext{cos}θ) = ext{cos}θ imes 0 - ext{sin}θ imes 1 = - ext{sin}θ.$

Hence, we have proven:

$rac{d}{dθ}( ext{cos}θ) = - ext{sin}θ.$

Given that θ is measured in radians, prove, from first principles, that d dθ (cosθ) = -sinθ You may assume the formula for cos(A ± B) and that as h → 0, sin(h)/h → 1 and cos(h) - 1/h → 0. - Edexcel - A-Level Maths Pure - Question 11 - 2018 - Paper 2

Worked Solution & Example Answer:Given that θ is measured in radians, prove, from first principles, that d dθ (cosθ) = -sinθ You may assume the formula for cos(A ± B) and that as h → 0, sin(h)/h → 1 and cos(h) - 1/h → 0. - Edexcel - A-Level Maths Pure - Question 11 - 2018 - Paper 2

Using the Definition of the Derivative

Simplifying the Expression

Using Known Limit Results

Join the A-Level students using SimpleStudy...