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Using small angle approximations, show that for small, non-zero values of $x$: $$\frac{x \tan 5x}{\cos 4x - 1} \approx A$$ where $A$ is a constant to be determined. - AQA - A-Level Maths: Pure - Question 4 - 2020 - Paper 2

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Using-small-angle-approximations,-show-that-for-small,-non-zero-values-of-$x$:--$$\frac{x-\tan-5x}{\cos-4x---1}-\approx-A$$--where-$A$-is-a-constant-to-be-determined.-AQA-A-Level Maths: Pure-Question 4-2020-Paper 2.png

Using small angle approximations, show that for small, non-zero values of $x$: $$\frac{x \tan 5x}{\cos 4x - 1} \approx A$$ where $A$ is a constant to be determined... show full transcript

Worked Solution & Example Answer:Using small angle approximations, show that for small, non-zero values of $x$: $$\frac{x \tan 5x}{\cos 4x - 1} \approx A$$ where $A$ is a constant to be determined. - AQA - A-Level Maths: Pure - Question 4 - 2020 - Paper 2

Step 1

Using small angle approximation for \( \tan 5x \)

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Answer

For small angles, we can use the approximation:

tanθθ\tan \theta \approx \theta

Therefore, for (\tan 5x):

tan5x5x\tan 5x \approx 5x

Step 2

Using small angle approximation for \( \cos 4x \)

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Answer

The cosine function can be approximated as:

cosθ1θ22\cos \theta \approx 1 - \frac{\theta^2}{2}

Thus for (\cos 4x):

cos4x1(4x)22=18x2\cos 4x \approx 1 - \frac{(4x)^2}{2} = 1 - 8x^2

Step 3

Substituting expressions into the original formula

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Answer

Substituting the approximations into the expression, we have:

xtan5xcos4x1x(5x)(18x2)1\frac{x \tan 5x}{\cos 4x - 1} \approx \frac{x (5x)}{(1 - 8x^2) - 1}

This simplifies to:

5x28x2=58\frac{5x^2}{-8x^2} = -\frac{5}{8}

Step 4

Determining \( A \)

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Answer

From the previous step, we find:

A=58A = -\frac{5}{8}

Thus, we have shown that:

xtan5xcos4x158\frac{x \tan 5x}{\cos 4x - 1} \approx -\frac{5}{8}

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