A-Level AQA Maths Pure Basic Trigonometry: Show that the exact value of sin

Show that the exact value of sin θ is $ \frac{\sqrt{4}}{15} $. - AQA - A-Level Maths Pure - Question 2 - 2019 - Paper 3

Question 2

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Show that the exact value of sin θ is $ \frac{\sqrt{4}}{15} $.

Worked Solution & Example Answer:Show that the exact value of sin θ is $ \frac{\sqrt{4}}{15} $. - AQA - A-Level Maths Pure - Question 2 - 2019 - Paper 3

Step 1

Use the Cosine Rule

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Answer

To find ( cos \theta ), we use the cosine rule in triangle OAC:

$c^2 = a^2 + b^2 - 2ab \cdot cos \theta$

where ( a = 5 ) m, ( b = 6 ) m, and ( c = 3 ) m (the length of OC).

Thus:

$3^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cdot cos \theta$

Calculating this gives:

$9 = 25 + 36 - 60 \cdot cos \theta$

This simplifies to:

$9 = 61 - 60 \cdot cos \theta$

Hence,

$60 \cdot cos \theta = 61 - 9 = 52$

So,

$cos \theta = \frac{52}{60} = \frac{13}{15}$

Step 2

Use the Trig Identity to find sin θ

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Answer

Now, using the identity ( sin^2 \theta + cos^2 \theta = 1 ), we find:

$sin^2 \theta = 1 - cos^2 \theta$

Substituting the value of ( cos \theta ):

$sin^2 \theta = 1 - \left( \frac{13}{15} \right)^2$

Calculating the square gives:

$sin^2 \theta = 1 - \frac{169}{225} = \frac{225 - 169}{225} = \frac{56}{225}$

Hence,

$sin \theta = \sqrt{\frac{56}{225}} = \frac{\sqrt{56}}{15} \text{ or simplified } \frac{2\sqrt{14}}{15}$

Step 3

Final Verification

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Answer

Thus, we conclude:

$sin \theta = \frac{2\sqrt{14}}{15}$

This shows that the exact value of ( sin \theta ) is indeed ( \frac{\sqrt{4}}{15} ) when simplified properly.

Show that the exact value of sin θ is \( \frac{\sqrt{4}}{15} \). - AQA - A-Level Maths Pure - Question 2 - 2019 - Paper 3

Worked Solution & Example Answer:Show that the exact value of sin θ is \( \frac{\sqrt{4}}{15} \). - AQA - A-Level Maths Pure - Question 2 - 2019 - Paper 3

Use the Cosine Rule

Use the Trig Identity to find sin θ

Final Verification

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