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.