Question

Find the value of sin 75 deg and sin 15 deg

Answer 1

sin 75 = cos 15 = root3 + 1 / 2root 2
sin 15 = root3 - 1 / 2root 2

Answer 2

$\bold{\huge{\underline{Solution-}}}$

$\bf \: Used \: formula - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: sin \: (A + B) = sinA \: cosB + cosA \: sinB}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: sin \: (A - B) = sinA \: cosB - cosA\: sinB}$

1.

$\bf \: sin \: 75 \degree \: = sin \: (45 \degree \: + 30 \degree) \: \\ \\ \boxed{\because \: 75\degree = 45 \degree + 30 \degree} \: \\ \\ \bf \: now \: using \: the \: sin(A + B) \:where \: A = 45 \degree \: B = 30 \degree \: \\ \\ \implies \bf \: sin \:45 \degree cos \: 30\degree + cos \: 45 \degree \: sin \: 30 \degree \: \\ \\ \implies \bf \: \frac{1}{ \sqrt{2} } . \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } . \frac{1}{2} \\ \\ \implies \bf \: \frac{ \sqrt{3} }{ \sqrt[2]{2} } + \frac{1}{ \sqrt[2]{2} } \\ \\ \implies \bf \: \frac{ \sqrt{3} + 1 }{ \sqrt[2]{2} }$

2.

$\bf \: sin \: 15 \degree \: = sin \: (45 \degree \: - 30 \degree) \: \\ \\ \boxed{\because \: 15\degree = 45 \degree - 30 \degree} \: \\ \\ \bf \: now \: using \: the \: sin(A - B) \:where \: A = 45 \degree \: B = 30 \degree \: \\ \\ \implies \bf \: sin \:45 \degree cos \: 30\degree - cos \: 45 \degree \: sin \: 30 \degree \: \\ \\ \implies \bf \: \frac{1}{ \sqrt{2} } . \frac{ \sqrt{3} }{2} - \frac{1}{ \sqrt{2} } . \frac{1}{2} \\ \\ \implies \bf \: \frac{ \sqrt{3} }{ \sqrt[2]{2} } - \frac{1}{ \sqrt[2]{2} } \\ \\ \implies \bf \: \frac{ \sqrt{3} - 1 }{ \sqrt[2]{2} }$