Question

if alpha is a root of 25 cos square theta + 5 cos theta - 12 = 0 and π/2 Less than alpha less than π then sin alpha =​

Answer 1

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$If \alpha \ is \ a\ root\ of\ 25 cos^2 \theta + 5 cos \theta - 12 = 0\\ and\ \displaystyle\frac{\pi}{2}< \alpha< \pi \ then\ find\ sin \alpha$

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$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Answer}}}}}{\bigstar}$

$\boxed{\red{\bold{Solving \ quadratic \ equation:}}}$

$\purple{\implies 25cos^2\theta + 5cos \theta -12 =0}$

$\purple{\implies}25 cos^2 \theta + 20 cos \theta - 15 cos \theta -12 =0$

$\purple{\implies}5 cos \theta(5 cos\theta +4) - 3 (5 cos \theta +4) = 0$

$\purple{\implies}(5 cos \theta + 4)(5 cos \theta -3)=0$

$\purple{\implies}cos \theta = \displaystyle \frac{-4}{5} \; \; or \; \; cos \theta = \frac{3}{5}$

$\blue{\alpha \ is \ the \ root \ of \ the \ equation ; \therefore \alpha = cos \theta}$

$\boxed{\red{\bold{As \ \displaystyle{\frac{\pi}{2}}< \alpha < \pi :}}}$

$\blue{\implies cos \theta \: is \: negative}$

$\blue{\implies}\therefore cos \theta = \displaystyle \frac{-4}{5}$

$\blue{\implies}\alpha = \displaystyle \frac{-4}{5}$

$\boxed{\red{\bold{By \: identity:}}}$

$\red{\implies}sin ^2 \theta + cos ^2 \theta = 1$

$\red{\implies}sin^2 \alpha + \alpha ^2 = 1$

$\red{\implies}sin^2 \alpha + \displaystyle \left( \frac{-4}{5} \right) ^2 = 1$

$\red{\implies}sin^2 \alpha + \displaystyle \frac{16}{25} = 1$

$\red{\implies}sin^2 \alpha = 1 - \displaystyle \frac{16}{25}$

$\red{\implies}sin^2 \alpha = \displaystyle \frac{9}{25}$

$\boxed{\red{\bold{Taking \ square \ root:}}}$

$\green{\implies sin \alpha = \displaystyle \sqrt{ \frac{9}{25}}}$

$\green {\implies}sin \alpha =\pm \displaystyle \frac{3}{5}$

$\boxed{\red{\bold{As \ \displaystyle \frac{\pi}{2}< \alpha <\pi}}}$

$\green{\boxed{\implies{\boxed{sin \alpha = \displaystyle \frac{3}{5} }}}}$

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