Question

1) If sin alpha = -4/5, and cos alpha>0, find tan alpha and sec alpha.

2) If cot x = -3/2 and sec x<0, find sin x and cos x.
3) If cos theta = 0.54, find sin(theta-pi/2).
4) If cot x = -0.18, find tan(x-pi/2).
Please explain/show steps, I'm very lost on how to do these!

Answer 1

1)

cos(α) > 0, is just another way to say cos(α) is positive, and therefore, the adjacent side is positive then.

now, on -4/5, keep in mind that the hypotenuse is just a radius unit, and therefore is never negative, so in -4/5, the negative has to be the -4.


`\bf sin(\alpha)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm√(c^2-b^2)=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(5^2-(-4)^2)=a\implies \pm 3=a\implies \stackrel{cos(\alpha)\ \textgreater \ 0}{+3=a}\\\\ -------------------------------`


`\bf tan(\alpha)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{adjacent}{3}}\qquad \qquad sec(\alpha)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{3}}`



2)


`\bf cot(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{opposite}{2}}\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√((-3)^2+2^2)\implies c=√(13)\\\\ -------------------------------`


`\bf sin(x)=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{√(13)}}\quad \stackrel{rationalizing~it}{\implies }\quad \cfrac{2}{√(13)}\cdot \cfrac{√(13)}{√(13)}\implies \cfrac{2√(13)}{13} \\\\\\ cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{√(13)}}\quad \stackrel{rationalizing~it}{\implies }\quad \cfrac{-3}{√(13)}\cdot \cfrac{√(13)}{√(13)}\implies -\cfrac{3√(13)}{13}`

hmm the sec(x) < 0, is just another way to say sec(x) is negative, and since the cosine is the reciprocal of the secant, then the cosine is also negative. Now in the fraction -3/2, which is the negative? since it could be -3/2 or 3/-2, well, anyhow, is the -3, because is the adjacent side, which is used by the cosine, that we know is negative.