Question 925540: How do I evaluate these this two sine and cosine problems. Difficult!?
Evaluate cos(a − b) if cos a = 4/5 tan a < 0, tan b = −√15 and cos b < 0
cos(a-b) =
and
Evaluate sin(x + y) if sin x = 3/5 sec x > 0, cos y = − (2√5)/5 and tan y < 0.
sin(x + y) =
Please give details
THANKS!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Evaluate cos(a − b) if cos a = 4/5 tan a < 0, tan b = −√15 and cos b < 0
cos(a-b) =
and
Evaluate sin(x + y) if sin x = 3/5 sec x > 0, cos y = − (2√5)/5 and tan y < 0.
sin(x + y) =
***
cos(a)=4/5
tan(a)<0
reference angle a is in quadrant IV where cos>0, sin<0
sin(a)=3/5 (working with a (3-4-5) reference right triangle in quadrant IV)
..
tan(b)=-√15/1
cos(b)>0
reference angle b is in quadrant IV where cos>0, sin<0
hypotenuse of reference right triangle=√(√15^2)+1^2)=√(16)=4
cos(b)=1/4
sin(b)=-√15/4
..
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
=4/5*1/4+3/5*√15/4
=4/20+3√15/20=4+3√15/20
..
check:
cos(a)=4/5
a≈323.13˚
tan(b)=-√15
b≈284.48
..
a-b≈323.13-284.48≈38.62˚
cos(a-b)≈cos(38.65)≈0.7809
exact value as computed=4+3√15/20≈0.7809
..
sinx=3/5
secx>0
reference angle x is in quadrant I where cos>0, sin>0
cosx=4/5 (working with a (3-4-5) reference right triangle in quadrant I)
..
cosy=-2√5/5
tany<0
reference angle y is in quadrant II where cos<0, sin>0
siny=√(1-cos^2(y))=√(1-20/25)=√(5/25)=√5/5
sin(x+y)=sinxcosy+cosxsiny=3/5*-2√5/5+4/5*√5/5=-6√5/25+4√5/25=-2√5/25
..
check:
sinx=3/5
x=36.87
cosy=-2√5/5
y=153.43
x+y=36.87+153.43=190.29
sin(x+y)=sin(190.30)≈-0.1788
exact value as computed=-2√5/25≈-0.1788
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