SOLUTION: Let sin A =12/13 with A in QII and sin B = − 15/17 with B in QIII. Find sin(A + B), cos(A + B), and tan(A + B).

Question 731793: Let
sin A =12/13
with A in QII and
sin B = − 15/17
with B in QIII. Find sin(A + B), cos(A + B), and tan(A + B).
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Let
sin A =12/13
with A in Q2 and
sin B = − 15/17
with B in Q3. Find sin(A + B), cos(A + B), and tan(A + B).
**
let o=opposite side
let a=adjacent side
let h=hypotenuse
..
sin A =12/13=o/h
o=12, h=13
a=-√h^2-o^2)=-√(13^2-12^2)=-√(169-144)=-√25=-5
cos A=a/h=-5/13
tan A=o/a=-12/5
..
sinB =-15/17=o/h
o=-15, h=17
a=-√h^2-o^2)=-√(17^2-15^2)=-√(289-225)=-√64=-8
cos B=a/h=-8/17
tan B=o/a=-15/-8=15/8
..
sin(A+B)=(sinA*cosB)+(cosA*sinB)
=(12/13)*(-8/17)+(-5/13*-15/17)
=-96/221+75/221
=-21/221
..
cos(A+B)=cosA*cosB)-(sinA*sinB)
=(-5/13)*(-8/17)-(12/13*-15/17)
=40/221+180/221
=220/221
..
tan(A+B)=(tanA+tanB)/(1-tanA*tanB
=(-12/5+15/8)/(1-(-12/5)(15/8))
=(-96/40+75/40)/(1+180/40
=(-21/40)/(220/40)
=-21/220
..
Check with calculator:
sinA=12/13 in Q2
A≈112.62�
sin B=-15/17 in Q3
B≈241.93�
A+B≈354.55�
reference angle=5.45� in Q4
..
sin(A+B)=sin(354.55)=sin(5.45) in Q4≈-0.0945...
-21/221=≈-0.0945...
..
cos(A+B)=cos(354.55)=cos(5.45) in Q4≈0.9954...
220-221≈0.9954...
..
tan(A+B)=tan(354.55)=tan(5.45) in Q4≈-0.0954...
-21/220≈-0.0954...

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SOLUTION: Let sin A =12/13 with A in QII and sin B = &#8722; 15/17 with B in QIII. Find sin(A + B), cos(A + B), and tan(A + B).

SOLUTION: Let sin A =12/13 with A in QII and sin B = − 15/17 with B in QIII. Find sin(A + B), cos(A + B), and tan(A + B).