Question

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

Answer 1

4√5/5

Explanation:

from

`(1)/( \ \cos( \alpha ) ) ?) = √(5)`

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

dividing through by

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

`{ \tan( \alpha ) }^(2) + 1 = (1)/(\cos( \alpha ) )`

`{ \tan( \alpha ) }^(2) = { √(5 ) }^(2) - 1`

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

`\tan \alpha = √(4 ) = 2`

but

`\tan( \alpha ) = ( \sin( \alpha ) )/( \cos( \alpha ) )`

`2 = \sin( \alpha ) * (1)/( \cos( \alpha ) )`

`2 = \sin( \alpha ) * √(5)`

`\sin( \alpha ) = (2)/( √(5) )`

`\sin( \alpha ) = (2 √(5) )/(5)`

Answer 2

`\bf sec(\theta )=√(5)\implies \cfrac{1}{cos(\theta )}=√(5)\implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{√(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√(c^2-a^2)=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}`

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