Find the exact value of the following under the given conditions a.cos(a-b) b.(a-b) c.tan(a-b)

Question

cos b=5/13, 3π/2<β<2π

csc=-5/4, π<β<3π/2

sum and difference identites

Answer 1

Hi Zoey C.,

With SO = Side Opposite, SA = Side Adjacent, and H = Hypotenuse:

sin(θ) = SO/H; cos(θ) = SA/H; tan(θ) = SO/SA; cot(θ) = SA/SO; sec(θ) = H/SA; csc(θ) = H/SO.

For cos(b) = 5/13 with (3π/2 < β < 2π) in quadrant IV. Therefore 5 = SA and 13 = H. This form a right triangle in quad. IV with a positive x value of 5 and a hypotenuse of 13. We can find SO from the Pythagorean theorem (H² = SO² + SA²), SO = ±√(13² - 5²) = ±12. And since this triangle is in quad. IV we choose -y or SO = -12.

We can do the same for csc(a) = -5/4 in the same quad. IV. H = 5 and SO = -4 (the hypotenuse is never negative and SO in quad IV corresponds to -y values), and SA = ±√(5² - [-4]²) = 3, (quad IV x values are +).

For a.)

cos(a - b) = cos(a)*cos(b) + sin(a)*sin(b), [subtraction formula]

cos(a - b) = (3/5)*(5/13) + (-4/5)*(-12/13)

cos(a - b) = 63/65

For b.) , a = sin^-1(-4/5) = -.93 rad and b = sin^-1(-12/13) = -1.18 rad, and (a - b) = .25 rad. No way to get an exact value. Is it supposed to be sin(a - b)?

For c.)

tan(a - b) = [tan(a) - tan(b)] / [1 + tan(a)*tan(b)]

tan(a - b) = [(-4/3) - (-12/5)] / [1 + (-4/3)*(-12/5)]

tan(a - b) = 16/63

I hope this helps, Joe.

P.S. Zoey C., now that you've seen how to do these maybe you can try your expertise at the others you have posted. Let me know if you need more help.