Find the exact value of sin(A-B), cos(A-B) and tan(A-B)

Question

Find the exact value of sin(A-B), cos(A-B) and tan(A-B)

When,

sin A = (2sqrt5) / 5 0<A< pi/2

cos B = 3/5 -pi/2<B<0

Answer 1

Calculating the exact value of sin(A - B), cos(A - B) and tan(A - B) requires the use of the trig angle subtraction identities:

sin(A – B) = sin(A)cos(B) – cos(A)sin(B)

cos(A – B) = cos(A)cos(B) + sin(A)sin(B)

tan(A - B) = (tan(A) - tan(B))/(1 + tan(A)tan(B))

and since we are only given sin(A) and cos(B), we must calculate the others from this information.

Given that sin(A) = (2√5)/5 and that the angle A is in Q1 and knowing that sin(θ) = opp/hyp, we can draw this sketch:

And we can solve for x using the Pythagorean Theorem:

x = √(5² - (2√5)²) = √(25 - 4•5) = √5

That means:

cos(A) = √5/5

tan(A) = 2√5/√5 = 2

Then, knowing cos(B) = 3/5, that angle B is in Q4, and that cos(θ) = adj/hyp we can draw a similar sketch of angle B:

And we can solve for y using the Pythagorean Theorem:

y = √(5² - 3²) = √(25 - 9) = √16 = 4 however this is going downwards so y = -4

That means:

sin(B) = -4/5

tan(B) = -4/3

We can now plug these values into the identities:

sin(A – B) = sin(A)cos(B) – cos(A)sin(B)

sin(A – B) = ((2√5)/5)(3/5) – (√5/5)(-4/5)

sin(A – B) = (6√5)/25 - (-4√5)/25

sin(A – B) = (6√5)/25 + (4√5)/25

sin(A – B) = (10√5)/25

sin(A – B) = (2√5)/5

(a strange answer but true)

You do the others just the same. Use the values we got to substitute into the identities and simplify.

Answer 2

cosA = √1 - 20/25) = √5/5; sinB = - √1 - 9/25 = - 4/5

sin(A - B) = sinAcosB - cosAsinB = 2√5/5·(3/5) - √5/5(- 4/5) = 6√5/25 + 4√5/25 = 10√5/25 = 2√5/5

cos(A - B) = cosAcosB + sinAsinB = √5/5·3/5 + 2√5/5·(- 4/5) = - 5√5/25 = - √5/5

tan(A - B) = sin(A - B)/cos(A - B) = (2√5/5)(- 5/√5) = - 2