Question

Please answer!!!

If cos0=-4/7, what are the values of sin0 and tan0?

Answer 1

let's first off notice something, the hypotenuse is never negative, since it's just a radius unit, so if the cosine is -(4/7), the hypotenuse of 7, is not the negative one, is the 4 above, so is really (-4)/7 in the fraction.

`\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-4}}{\stackrel{hypotenuse}{7}}~\hspace{5em}\textit{let's find they \underline{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√(7^2-(-4)^2)=b\implies \pm√(49-16)=b\implies \boxed{\pm√(33)=b} \\\\[-0.35em] ~\dotfill\\\\ sin(\theta )=\cfrac{\stackrel{opposite}{\pm √(33)}}{\stackrel{hypotenuse}{7}}~\hspace{7em}tan(\theta )=\cfrac{\stackrel{opposite}{\pm √(33)}}{\stackrel{adjacent}{-4}}\implies tan(\theta )=\cfrac{\stackrel{opposite}{\mp √(33)}}{\stackrel{adjacent}{4}}`