Question

3Sin thita -4Cos thita ki miximum value

Answer 1

Given:

3Sin thita -4Cos thita ki miximum value

Answer:

$\sqrt{2}$

Solution:

$\sin0 + \cos0 \\ = \sqrt{2} ( \frac{ \sin0 }{ \sqrt{2} } + \frac{ \cos0 }{ \sqrt{2} } \\ = \sqrt{2} ( \sin(0 + \frac{n}{4} )) \\ max \: value \: of \: sin \: (0 + \frac{n}{4} ) = 1 \\ min \: value \: of \: \sin(0 + \frac{n}{4} ) = \sqrt{2} \\ max \: value \: of \: \sin \: 0 + \cos0 = \sqrt{2}$