Question

If cos of theta = -4/9 with theta in quadrant 2, find sin of theta.

Answer 1

what are the signs for cosine and sine in the 2nd quadrant anyway?

both positive, both negative
cosine positive and sine negative or viceversa?

Answer 2

sines positive cosines negitive

Answer 3

\[\sin \left(\cos ^{-1}\left(-\frac{4}{9}\right)\right)=\frac{\sqrt{65}}{9} \]

Answer 4

\(\bf cos(\theta)=-\cfrac{4}{9}\to \cfrac{adjacent\ side}{hypotenuse}\to \cfrac{a=-4}{c=9}\) \(\bf c^2=a^2-{\color{blue}{ b}}^2\implies \pm \sqrt{c^2-a^2}={\color{blue}{ b}}
\\ \quad \\
\textit{we know that sine is positive, so we use the + of the root }
\\ \quad \\
+\sqrt{c^2-a^2}={\color{blue}{ b}}\qquad recall\implies sin(\theta)=\cfrac{opposite\ side}{hypotenuse}\to \cfrac{{\color{blue}{ b}}}{c}\)

Answer 5

so the answer is sqrt(97)/9?

Answer 6

hmm

Answer 7

or it is negative because sin is in quadrant 2 so -sqrt(97)/9 ???

Answer 8

\(\bf c^2=a^2-{\color{blue}{ b}}^2\implies \pm \sqrt{c^2-a^2}={\color{blue}{ b}}\implies +\sqrt{(9)^2-(-4)^2}={\color{blue}{ b}}
\\ \quad \\
\sqrt{81-16}={\color{blue}{ b}}\qquad sin(\theta)=\cfrac{\sqrt{81-16}}{9}\)

Answer 9

"sines positive cosines negitive" <--- sine is postive, cosine is negative

Answer 10

oooooh so sqrt(65)/9

Answer 11

yeap

Answer 12

ok thanks

Answer 13

yw