Question

3 COSA + 4 Sin A=4
Find sin A​

Answer 1

$\Large\frak{\underline{\underline{Correct \: Question:}}}$

$\normalsize\sf\ If \: 3cosA + 4sinA = 4, \: then \: find \\ \normalsize\sf\ sinA$

$\Large\frak{\underline{\underline{Answer:}}}$

$\normalsize\bullet\:\sf\ Let \: sinA = y$

$\normalsize\bullet\:\sf\ cosA= \sqrt{1 - sin^2A}$

$\normalsize\qquad\sf\ = \sqrt{1 - y^2}$

$\rule{170}2$

$\underline{\bigstar\:\sf{Block \: the \: values \: of \: sinA \: \& \: and \: cosA:}}$

$\normalsize\ : \implies\sf\ 3cosA + 4sinA = 4$

$\normalsize\ : \implies\sf\ 3cosA = 4 - 4sinA$

$\normalsize\ : \implies\sf\ 3cosA = 4(1 - sinA)$

$\normalsize\ : \implies\sf\ 3(\sqrt{1 - y^2}) = 4(1 - (y) )$

$\scriptsize\sf{\quad\dag\ Squaring \: both \; sides}$

$\normalsize\ : \implies\sf\ \left[3(\sqrt{1 - y^2}) \right]^2 = \left[4(1 - (y) ) \right]^2$

$\normalsize\ : \implies\sf\ 9(1 - y^2) = 16(1 - (y)^2 )$

$\normalsize\ : \implies\sf\ 9 - 9y^2 = 16(1 + y^2 - 2y)$

$\normalsize\ : \implies\sf\ 9 - 9y^2 = 16 + 16y^2 - 32y$

$\normalsize\ : \implies\sf\ 16 + 16y^2 - 32y - 9 + 9y^2 = 0$

$\normalsize\ : \implies\sf\ 25y^2 - 32y - 7 = 0$

$\rule{150}1$

$\scriptsize\sf{\quad\dag\ Using \: Quadratic \: formula}$

$\normalsize\dashrightarrow\sf\ y = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$\normalsize\dashrightarrow\sf\ y = \frac{-(-32) \pm \sqrt{(32)^2 - 4 \times\ 25 \times\ 7} }{2 \times\ 25}$

$\normalsize\dashrightarrow\sf\ y = \frac{32 \pm \sqrt{1024 - 700} }{50}$

$\normalsize\dashrightarrow\sf\ y = \frac{32 \pm \sqrt{324} }{50}$

$\normalsize\dashrightarrow\sf\ y = \frac{32</p><p>\pm 18 }{50}$

$\boxed{\begin{tabular}{ c | c } </p><p>\bf{y = +ve} & \bf{y = -ve}\\ \cline{1-2}\\ \sf{ y = \frac{32 + 18}{50} } & \sf{ y = \frac{32 - 18}{50} } \\ \\ \sf{ y = \frac{50}{50} } & \sf{ y = \frac{14}{50} } \\ \\ \sf{ y = 1 } & \sf{ y = \frac{7}{25} } \end{tabular}}$

$\therefore\:\underline{\textsf{Hence, \: the \: value \: of \: sinA(y) \: is \: }{\bf{ 1 \: or \: \frac{7}{25}}}}$