Question

What is sin θ when sec θ = square root of 5? Rationalize the denominator if necessary.

Answer 1

Answer:

4√5/5

Step-by-step explanation:

from

$\frac{1}{ \ \cos( \alpha ) ) ?} = \sqrt{5}$

${ \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1$

dividing through by

${ \cos( \alpha ) }^{2}$

${ \tan( \alpha ) }^{2} + 1 = \frac{1}{\cos( \alpha ) }$

${ \tan( \alpha ) }^{2} = { \sqrt{5 } }^{2} - 1$

${ \tan( \alpha ) }^{2} = 5 - 4 = 4$

$\tan \alpha = \sqrt{4 } = 2$

but

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

$2 = \sin( \alpha ) \times \frac{1}{ \cos( \alpha ) }$

$2 = \sin( \alpha ) \times \sqrt{5}$

$\sin( \alpha ) = \frac{2}{ \sqrt{5} }$

$\sin( \alpha ) = \frac{2 \sqrt{5} }{5}$

Answer 2

$\bf sec(\theta )=\sqrt{5}\implies \cfrac{1}{cos(\theta )}=\sqrt{5}\implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{5}}}=cos(\theta )~\hfill \leftarrow \stackrel{\textit{let's find the}}{\textit{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{(\sqrt{5})^2-1^2}=b\implies \pm\sqrt{5-1}=b\implies \pm\sqrt{4}=b\implies \pm 2=b \\\\[-0.35em] ~\dotfill\\\\ sin(\theta )=\pm\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{\sqrt{5}}}\implies \stackrel{\textit{rationalizing the denominator}}{\pm\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \pm\cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \pm\cfrac{2\sqrt{5}}{5}}$