Question

given sin theta =a/b then find the value of cos theta​

Answer 1

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Answer 2

It is given that the algebraic value of sin$\theta$ is $\dfrac{a}{b}$

$\implies \sin \theta = \dfrac{a}{b}$


Square on both sides of the equation :

$\implies \sin {}^{2} \theta = \dfrac{a {}^{2} }{ {b}^{2} }$


From the properties of trigonometry, we know
• sin^2 A = 1 - cos^2 A


Thus,
$\implies 1 - \cos {}^{2} \theta = \dfrac{ {a}^{2} }{ {b}^{2} } \\ \\ \\ \implies \cos {}^{2} \theta = 1 - \dfrac{ {a}^{2} }{b {}^{2} } \\ \\ \\ \implies { \cos}^{2} \theta = \dfrac{ {b}^{2} - {a}^{2} }{b {}^{2} } \\ \\ \\ \implies \cos {}^{} \theta = \sqrt{ \dfrac{ {b}^{2} - {a}^{2} }{ {b}^{2} } } \\ \\ \\ \implies\cos\theta = \dfrac{ \sqrt{b {}^{2} - {a}^{2} } }{b}$

Hence the algebraic value of cos $\theta$, when sin$\theta$ is $\dfrac{a}{b}$, is $\dfrac{ \sqrt{b {}^{2} - {a}^{2} } }{b}$