Question

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$\huge \: hlo \: guys \\ \\ \huge \: gm \\ \\ \small \: find \: the \: value \: of \: sin \: 15°$


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Answer 1

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$\huge \: hlo \: guys \\ \\ \huge \: gm$

Sin 15°= Sin(45–30)°

W.k.t, Sin(A-B) = SinACosB - CosASinB

Therefore, Sin(45–30)°= Sin45°Cos30° - Cos45°Sin30°

W.k.t, sin45°= cos45°=1/√2, sin30°= 1/2, cos30°=√3/2

Therefore, Sin(45–30)°= (1/√2)(√3/2)-(1/√2)(1/2)

= (√3/2√2)-(1/2√2)

= (√3–1)/2√2

I hope it help u ☺.

Answer 2

$\small \bf \red{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \red{ \huge \: solution} \\ \\\small \bf \red{sin \: 15} \\ \\ \small \bf \red{as \: we \: can \: write} \\ \\ \small \bf \red{sin \: 15 = sin(60 - 45)} \\ \\ \small \bf \red{sin(60 - 45)} \\ \\ \small \bf \red{using \: this \: formula} \\ \\ \small \bf \red{sin(a + b) = sin \: a \: \times cos \: b - cos \: a \times sin \: b} \\ \\ \small \bf \red{sin \: (60 - 45) = sin \: 60 \times cos \: 45 - cos \: 60 \times sin \: 45} \\ \\ \small \bf \red{sin \: 15 = \frac{ \sqrt{3} }{2} \times \frac{1}{ \sqrt{2} } - \frac{1}{2} \times \frac{1}{ \sqrt{2} } } \\ \\ \small \bf \red{sin \: 15 = \frac{ \sqrt{3} }{2 \sqrt{2} } - \frac{1}{2 \sqrt{2} } } \\ \\ \small \bf \red{sin \: 15 = \frac{ \sqrt{3} - 1}{2 \sqrt{2} } }$