Question

If sin θ = 4/5 and cos θ is in quadrant II, then cos θ =

Answer 1

In quadrant II, the cosine function is negative. We can use the Pythagorean identity to find cos θ. Since cosine is negative, the value of cos θ is -3/5.

Step-by-step explanation:

In quadrant II, the cosine function is negative. Since sin θ = 4/5, we can use the Pythagorean identity to find cos θ:

sin2θ + cos2θ = 1

Substituting the value of sin θ:

(4/5)2 + cos2θ = 1

Simplifying the equation:

16/25 + cos2θ = 1

cos2θ = 1 - 16/25

cos2θ = 9/25

Taking the square root of both sides:

cos θ = ± 3/5

Since cosine is negative in quadrant II, we take the negative value:

cos θ = -3/5

Answer 2

keeping in mind that in the II quadrant, since is positive and the cosine is negative, then,


`\bf sin(\theta )=\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}`


`\bf \pm√(5^2-4^2)=a\implies \pm√(25-16)=a\implies \pm√(9)=a \\\\\\ \pm 3=a\implies \stackrel{\textit{II~quadrant}}{-3} \\\\\\ cos(\theta )=\cfrac{adjacent}{hypotenuse}\qquad \qquad \qquad cos(\theta )=\cfrac{-3}{5}`