Asked by Ju
If sinx= 5/13, and x is a positive acute angle, find sin (x + 3pi/2)
Answers
Answered by
Anonymous
(All angles in radians)
sin(x)=5/13
cos(x)= + OR - sqroot[1-sin^2(x)]
In this case x is positive so:
cos(x)=sqroot[1-sin^2(x)]
cos(x)=sqroot(1 - 25/169)
cos(x)=sqroot(169/169 - 25/169)
cos(x)=sqroot(144/169)
cos(x)=12/13
sin(3pi/2)= -1
cos(3pi/2)=0
sin(alpha+beta)=sin(alpha)*cos(beta)+cos(alpha)*sin(beta)
sin(x+3pi/2)=sin(x)*cos(x+3pi/2)+cos(x)*sin(x+3pi/2)=
(5/13)*0+(12/13)*(-)1=0 - 12/13= -12/13
sin(x+3pi/2)= -12/13
sin(x)=5/13
cos(x)= + OR - sqroot[1-sin^2(x)]
In this case x is positive so:
cos(x)=sqroot[1-sin^2(x)]
cos(x)=sqroot(1 - 25/169)
cos(x)=sqroot(169/169 - 25/169)
cos(x)=sqroot(144/169)
cos(x)=12/13
sin(3pi/2)= -1
cos(3pi/2)=0
sin(alpha+beta)=sin(alpha)*cos(beta)+cos(alpha)*sin(beta)
sin(x+3pi/2)=sin(x)*cos(x+3pi/2)+cos(x)*sin(x+3pi/2)=
(5/13)*0+(12/13)*(-)1=0 - 12/13= -12/13
sin(x+3pi/2)= -12/13
Answered by
Anonymous
(x+3pi/2) radians = 270°
Answered by
Anonymous
Sorry mistake.
3pi/2 radians = 270°
3pi/2 radians = 270°
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