Question

If A= 45° , Verify that Sin3A = 3sinA – 4sin3 A. pls solve neatly​

Answer 1

To Prove :

$\sf \implies \sin3A = 3 \sin A - 4 { \sin}^{3} A$

When A = 45°

Proof :

$\sf\implies\sin3A = 3 \sin A - 4 { \sin}^{3} A\\ \\\sf\implies\sin3 \times 45 = 3 \sin45 - 4 { \sin}^{3}45 \\ \\ \sf\implies\sin135 = 3 \times \frac{1}{ \sqrt{2} } - 4 \times ( { \frac{1}{ \sqrt{2} }) }^{3} \\ \\\sf\implies \cos45 = \frac{3}{ \sqrt{2} } - 4 \times \frac{1}{ \sqrt{8} } \\ \\ \sf\implies \frac{1}{ \sqrt{2} } = \frac{3}{ \sqrt{2} } - \frac{4}{ \sqrt{8} } \\ \\\sf\implies \frac{1}{ \sqrt{2} } = - \frac{6 - 4}{ \sqrt{8} } \\ \\ \sf\implies \frac{1}{ \sqrt{2} } = \frac{2}{ \sqrt{8} } \\ \\\sf\implies \frac{1}{ \sqrt{2} } = \frac{2}{2 \sqrt{2} } \\ \\\sf\implies \frac{1}{ \sqrt{2} } = \frac{1}{ \sqrt{2} }$

LHS = RHS

Hence Proved

Answer 2

$\large\red{\underline{\boxed{\textbf{Brainliest\:Please}}}}$