Question

find the exact values of sin(a+b) and Tan (a+b) given that sina =(4/5) and cosb = (-12/13) where a is in quadrant I and b is in quadrant II

Answer 1

Do you know the equation for sin(a+b)?

Answer 2

sin(a+b) = sinacosb + cosasinb

Answer 3

sin(a+b) = sin(a)cos(b)+cos(a)sin(b)

tan(a+b) =

\[\frac{ \sin(a)\cos(b) }{ \cos(a)\cos(b)-\sin(a)\sin(b) }+ \frac{ \cos(a)\sin(b) }{ \cos(a)\cos(b)-\sin(a)\sin(b)}\]

Answer 4

\(\bf sin(a)=\cfrac{4}{5}\implies \cfrac{opposite}{hypotenuse}\implies \cfrac{b=4}{c=5}\qquad Quadrant\ I
\\ \quad \\
cos(b)=-\cfrac{12}{13}\implies \cfrac{adjacent}{hypotenuse}\implies\cfrac{a=-12}{c=13}\qquad Quadrant\ II\)

Answer 5

In order to find angle a and b, you can use arc trig functions, or solve the triangles given above

Answer 6

or just use the pythagorean theorem to find the missing side

once you get that, you'd have the cos(a) and also the sin(b)

recall that the SUM trig identities are sine and cosine based, then just plug them in

Answer 7

Thank you so much