If sinx= 5/13, and x is a positive acute angle, find sin (x + 3pi/2)
(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
(x+3pi/2) radians = 270°
Sorry mistake.
3pi/2 radians = 270°
To find sin(x + 3π/2), we can use the angle addition formula for sine, which states:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
In this case, we need to find sin(x + 3π/2), where sin(x) = 5/13.
Let's break it down step by step:
1. We know that sin(x) = 5/13. Since x is a positive acute angle, it means x is in the first quadrant.
2. To find cos(x), we can use the Pythagorean identity: sin^2(x) + cos^2(x) = 1. Plugging in sin(x) = 5/13, we can solve for cos(x):
(5/13)^2 + cos^2(x) = 1
25/169 + cos^2(x) = 1
cos^2(x) = 1 - 25/169
cos^2(x) = 144/169
cos(x) = sqrt(144/169) = 12/13
3. Now, we need to find sin(3π/2). Since 3π/2 is in the third quadrant, we know that sin(3π/2) = -1.
4. Putting it all together, we can use the angle addition formula:
sin(x + 3π/2) = sin(x)cos(3π/2) + cos(x)sin(3π/2)
Substituting sin(x) = 5/13, cos(x) = 12/13, and sin(3π/2) = -1:
sin(x + 3π/2) = (5/13)(0) + (12/13)(-1)
sin(x + 3π/2) = 0 - 12/13
sin(x + 3π/2) = -12/13
Therefore, sin(x + 3π/2) = -12/13.