If sin alpha = 1/sqrt{5} and sin beta = 3/5,t | vedclass

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(A) Given $\sin \alpha = 1/\sqrt{5}$,then $\cos \alpha = \sqrt{1 - (1/\sqrt{5})^2} = 2/\sqrt{5}$.Given $\sin \beta = 3/5$,then $\cos \beta = \sqrt{1 -

Vedclass · Accepted Answer

$[0, \pi/4]$

Vedclass · Answer

(A) Given $\sin \alpha = 1/\sqrt{5}$,then $\cos \alpha = \sqrt{1 - (1/\sqrt{5})^2} = 2/\sqrt{5}$.Given $\sin \beta = 3/5$,then $\cos \beta = \sqrt{1 - (3/5)^2} = 4/5$.Using the formula $\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha$:$\sin(\beta - \alpha) = (3/5)(2/\sqrt{5}) - (4/5)(1/\sqrt{5}) = 6/(5\sqrt{5}) - 4/(5\sqrt{5}) = 2/(5\sqrt{5})$.Since $5\sqrt{5} \approx 11.18$,$\sin(\beta - \alpha) \approx 2/11.18 \approx 0.1789$.We know $\sin(0) = 0$ and $\sin(\pi/4) = 1/\sqrt{2} \approx 0.7071$.Since $0 Thus,$\beta - \alpha$ lies in the interval $[0, \pi/4]$.

If $\sin \alpha = 1/\sqrt{5}$ and $\sin \beta = 3/5$,then $\beta - \alpha$ lies in the interval

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