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Exercise 3.1 · Q1

Q.Find the value of sin⁡15°\sin 15°

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✓ Free question

Step 1: Write 15°=45°−30°15°=45°-30°.

Step 2: Apply sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B)=\sin A\cos B-\cos A\sin B with A=45°,B=30°A=45°,B=30°: sin⁡15°=sin⁡45°cos⁡30°−cos⁡45°sin⁡30°\sin15°=\sin45°\cos30°-\cos45°\sin30°.

Step 3: Substitute the standard values sin⁡45°=cos⁡45°=12\sin45°=\cos45°=\dfrac{1}{\sqrt2}, cos⁡30°=32\cos30°=\dfrac{\sqrt3}{2}, sin⁡30°=12\sin30°=\dfrac{1}{2}: sin⁡15°=12⋅32−12⋅12=3−122\sin15°=\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2}\cdot\dfrac{1}{2}=\dfrac{\sqrt3-1}{2\sqrt2}.

Step 4: Rationalise: 3−122=6−24\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac{\sqrt6-\sqrt2}{4}.

✓Final answer

sin⁡15°=6−24\sin15°=\dfrac{\sqrt6-\sqrt2}{4}

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