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Given that tan θ = p, where p is a constant, p ≠ ±1 use standard trigonometric identities, to find in terms of p, (a) tan 2θ (b) cos θ (c) cot(θ − 45°) Write each answer in its simplest form. - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 3

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Given-that--tan-θ-=-p,-where-p-is-a-constant,-p-≠-±1--use-standard-trigonometric-identities,-to-find-in-terms-of-p,--(a)-tan-2θ--(b)-cos-θ--(c)-cot(θ-−-45°)--Write-each-answer-in-its-simplest-form.-Edexcel-A-Level Maths Pure-Question 3-2015-Paper 3.png

Given that tan θ = p, where p is a constant, p ≠ ±1 use standard trigonometric identities, to find in terms of p, (a) tan 2θ (b) cos θ (c) cot(θ − 45°) Write e... show full transcript

Worked Solution & Example Answer:Given that tan θ = p, where p is a constant, p ≠ ±1 use standard trigonometric identities, to find in terms of p, (a) tan 2θ (b) cos θ (c) cot(θ − 45°) Write each answer in its simplest form. - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 3

Step 1

tan 2θ

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Answer

Using the double angle formula for tangent, we have:

tan2θ=2tanθ1tan2θ\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}

Substituting (\tan \theta = p):

tan2θ=2p1p2\tan 2\theta = \frac{2p}{1 - p^2}

Thus, the answer in its simplest form is:

tan2θ=2p1p2\tan 2\theta = \frac{2p}{1 - p^2}

Step 2

cos θ

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Answer

To find (\cos \theta), we use the identity:

tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}

This leads to:

tan2θ+1=sec2θ\tan^2 \theta + 1 = \sec^2 \theta

Substituting (\tan \theta = p) gives:

p2+1=sec2θp^2 + 1 = \sec^2 \theta

Thus, we have:

sec2θ=1+p2\sec^2 \theta = 1 + p^2

Therefore:

cosθ=11+p2\cos \theta = \frac{1}{\sqrt{1 + p^2}}

Step 3

cot(θ − 45°)

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Answer

Using the identity for cotangent:

cot(θ45°)=cotθ+11cotθ\cot(\theta - 45°) = \frac{\cot \theta + 1}{1 - \cot \theta}

Knowing (\cot \theta = \frac{1}{\tan \theta} = \frac{1}{p}), we substitute:

cot(θ45°)=1p+111p\cot(\theta - 45°) = \frac{\frac{1}{p} + 1}{1 - \frac{1}{p}}

Simplifying this:

=1+ppp1p=1+pp1= \frac{\frac{1 + p}{p}}{\frac{p - 1}{p}} = \frac{1 + p}{p - 1}

Thus, the answer is:

cot(θ45°)=1+pp1\cot(\theta - 45°) = \frac{1 + p}{p - 1}

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