Question

Given sin u = .345 and cos v =.804 find tan(u+v)

Answer 1

\(\bf tan(\alpha+\beta)=\cfrac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)tan(\beta)}\qquad recall\to {\color{brown}{ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}}}
\)

Answer 2

hmm

Answer 3

yes I recall. question. I use cos u = sqrt 1-sin^2u to obtain cos u. is this correct?

Answer 4

I am really trying to understand. I do not want an answer just explanation.

Answer 5

yes... you could use that, yes

Answer 6

and then cos v = sqrt 1-sin^2v to obtain cos v

Answer 7

\(\bf sin^2(\theta)+cos^2(\theta)=1\implies cos(\theta)=\sqrt{1-sin^2(\theta)}\)

yeap

Answer 8

why do you switch it to 1-tan(a)tan(b)?

Answer 9

hmmmm o hh that's just the identity for tan(u+v) using tangent values

Answer 10

but I gather that may be a longer way

Answer 11

I see.

Answer 12

well.... actaully either way ... coiuld be just as long

because using the sin(a+b) and cos(a+b) you'd still have to expand anyhow

Answer 13

\[\frac{ \sin u }{ \cos u } - \frac{ \sin v }{ \cos v }\]

Answer 14

is in my numerator

Answer 15

?

Answer 16

\(\bf tan(u+v)\implies \cfrac{sin(u+v)}{cos(u+v)}\implies \cfrac{sin(u)cos(v)+cos(u)sin(v)}{cos(u)cos(v)-sin(u)sin(v)}\)