Question

Given tanA=5 and sinB=2/3
I need to find:
sin(A-B)
cos(A-B)
tan(A-B)

Answer 1

so.... what's stopping from getting sin(A-B)?

Answer 2

Because I dont know how to find Acos and Bcos

Answer 3

ahhh, well... ok... let's us notice something

Answer 4

Ya do you know how to find that with the given information?

Answer 5

\(\bf tan(A) = 5 \\ \quad \\ \quad \\
tan(\theta) = \cfrac{b}{a}\quad thus\\ \quad \\
tan(A) = 5\implies tan(A) = \cfrac{5}{1}\implies \cfrac{b}{a}\\ \quad \\
c^2 = a^2+b^2\implies c = \sqrt{a^2+b^2}\\ \quad \\
\textit{keep in mind that }\quad cos(A) = \cfrac{a}{c}\)

Answer 6

hmm... Im still not quite understanding.

Answer 7

the idea being that tan(A) = 5, or 5/1, keeping in mind the tangent identity of \(\bf tan(\theta) = \cfrac{\textit{opposite side}}{\textit{adjacent side}}\implies \cfrac{b}{a}\)

Answer 8

Okay I got that.. The what do I do?

Answer 9

you find the "c" side of angle A, once you find "c"
you find the cosine, because cos(A) = a/c

Answer 10

Okay so cos(A)=1/\[\sqrt{26}\]?

Answer 11

\[1/\sqrt{26}\]

Answer 12

\(\bf cos(A) = \cfrac{1}{\sqrt{26}}\) then

Answer 13

Okay so my last question is how would I find tanB?

Answer 14

the 2nd one is a bit ambiguous, because the angle could be on either the 1st or 2nd Quadrants, but anyhow

\(\bf sin(B) = \cfrac{2}{3}\implies \cfrac{b}{c}\\ \quad \\
c^2 = a^2 + b^2 \implies \sqrt{c^2-b^2}=a\\ \quad \\
\textit{keep in mind that }\quad cos(B) = \cfrac{a}{c}\)

Answer 15

the pythagorean theorem, doesn't tell us what the value should be, negative or minus, so
\(\bf a = \pm \sqrt{3^2-2^4}\implies a = \pm \sqrt{5}\)

Answer 16

negative or positive I meant

Answer 17

Okay thank you I have the rest of the equation figured out

Answer 18

yw

Answer 19

hmm 2^4? hehe, anyhow, I meant \(\bf a = \pm \sqrt{3^2-2^2}\implies a = \pm \sqrt{5}\)