If ⁡sin x=7/25, and 0<x<π2, what is cos(x−3π/2)?

Question

If ⁡sin x=7/25, and 0<x<π2, what is cos(x−3π/2)?

Answer 1

sin x = 7/25 = .28 arcsin.28 = 17 degrees or 163 degrees in the 1st or 2nd quadrant. Otherwise sin x <0 cos x = 24/25 or -24/25 in 1st or 2nd quadrant. 25 squared = 24 squared plus 7 squared

3pi/2 = 270 degrees. 0<x<360 degrees, but given it's sine>0 we know 0<x<180

x-270 will be in the 2nd or 4th quadrant, 17-270=-253 degrees=107 degrees or 163-270=--103 = 270-13=257 degrees. cos107 = 0.292 cos257=-0.225. But must be some rounding errors, as actual answer is 0.28. still close, within .055.

cos(A-B)=cosAcosB+SinAsinB In this problem sinA=7/25=.28

cos(x-270)=cos.28cos270+sin.28sine270 = 25/27(0)+7/25(1) = 7/25= .28

Answer 2

Hello Alice!

We can use:

cos(x−3π/2) = cosx*cos(3π/2) + sin(x)*sin(3π/2) = cosx*0 + sin(x)*1 = sin(x) = 7/25

Answer 3

let T = 3*pi/2 = 270

sin x = 7/25

cos(T) = 0 while sin(T) = -1

cos(x - 3*pi/2) = cos(x) cos (3*pi/2) + sin x * sin (3*pi/2) <--- angle subtraction formula for cosine

= cos(x) * 0 + (7/25) * -1 <-- plugs in the given values

= -7/25

= -0.28

As an APPROXIMATION....

sin x = 7/25, so

opposite = 7, hypotenuse = 25

Then adjacent = 24 by pythagorean

then x = inverse_sine(7/25) = 16.26 + degrees

x- 3*pi/2 = 16.26+ - 270 = -253.73+ = 106.26+

the cosine of this angle is slightly more than -0.3

which agrees with the previous value