If tan x= 3/4, and 0<x<π/2, what is cos(x−3π/2)?

Question

If tan x= 3/4, and 0<x<π/2, what is cos(x−3π/2)?

Answer 1

My explanation requires a re-write of the original problem:

If tan A = 3/4, and 0 < A < π/2, what is cos(A−3π/2)?

Since tan A = 3 / 4, then cos A = 4 / 5 and sin A = 3 / 5. To see this, draw a triangle on a coordinate plane as follows: From the origin, draw a line to the point (4,3). Draw a line from the point to the x-axis at x=4. The result is a 3-4-5 triangle, and angle A is the angle that the hypotenuse makes with the x-axis.

Remember the hypotenuse is the square root of the sum of the squares of the other two sides:

3² + 4² = 9 + 16 = 25 = 5²

Also, remember that cos A is the Adjacent side (4) divided by the hypotenuse (5), so cos A = 4/5

Finally, remember that sin A is the Opposite side (3) divided by the hypotenuse (5), so sin A = 3/5.

Now employ the Addition Formula for Cosine:

cos(A - B) = cos(A)*cos(B) + sin(A)*sin(B)

Since B = 3π/2, cos(B) = 0 and sin(B)= -1.

cos(A - B) = (4/5)*(0) + (3/5)*(-1) = -3/5, ans.

Answer 2

tan x = 3/4 --> opposite = 3 and adjacent = 4 --> hypotenuse = 5 by pythagorean

So then

tan x = 3/4 --> sin x = 3/5 and cos x = 4/5

Let T = 3*pi/2 = 270 degrees

Then

cos T = 0

sin T = -1

cos(x - T) = cosX cos T + sin x sin T

= cos X * 0 + (3/4)(-1)

= -3/4