Question

If sin(A-B) =sinA CosB - CosA SinB, find the value of sin 15°​

Answer 1

( √3 - 1 ) / 2√2

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$\orange{\bigstar}$  Given  $\green{\bigstar}$

sin(A-B) = sinA cosB - cosA sinB

$\orange{\bigstar}$  To Find  $\green{\bigstar}$

Value of sin 15°

$\orange{\bigstar}$  Solution  $\green{\bigstar}$

15 can be written as ( 60 - 45 )

$\to \bf sin(15)\\\\\to \rm sin(60-45)\\\\\to \rm sin(60).cos(45)-cos(60).sin(45)\\\\ \to \rm \dfrac{\sqrt{3}}{2}.\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}.\dfrac{1}{\sqrt{2}}\\\\\to \rm \dfrac{\sqrt{3}}{2\sqrt{2}}-\dfrac{1}{2\sqrt{2}}\\\\\to \bf \dfrac{\sqrt{3}-1}{2\sqrt{2}}$

So ,

$\bf \sin15^{\circ}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\ \; \pink{\bigstar}$

$\orange{\bigstar}$  Trigonometric Value's  $\green{\bigstar}$

$\bullet\ \; \rm sin\ 45=\dfrac{1}{\sqrt{2}}\\\\\bullet\ \; \rm sin\ 60=\dfrac{\sqrt{3}}{2}\\\\\bullet\ \; \rm cos\ 45=\dfrac{1}{\sqrt{2}}\\\\\bullet\ \; \rm cos\ 60=\dfrac{1}{2}$

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