Question

If sinB =1/2, prove that 3cosB – 4cos³B=0.

Answer 1

Method 1
$\sin(B) = \frac{1}{2} \\ \cos(B) = \sqrt{1 - { \sin }^{2}B } = \sqrt{1 - \frac{1}{4} } = \frac{ \sqrt{3} }{2} \\ 3 \cos(B) - 4 { \cos}^{3} B = 3 \times \frac{ \sqrt{3} }{2} - 4 \times {( \frac{ \sqrt{3} }{2} )}^{3} \\ = \frac{3 \sqrt{3} }{2} - \frac{3 \sqrt{3} }{2} = 0$

Method 2
$\sin(B) = \frac{1}{2}. \: \: \: \: so \: \: B = {30}^{0} \\ 3 \cos(B) - 4 { \cos }^{3} B = \cos(3B) \\ = \cos(3 \times 30) = \cos( {90}^{0} ) = 0$