Find the exact value of each of the following if 0≤x≤π/2 and 0≤y≤π/2: a) sin(x-y) if cos x =3/5 and sin y=24/25 b) tan(x+y) if cot x =6/5 and sec y=3/2

Answer 1

To do these problems, you need to know:

1) SOH CAH TOA - With this first problem, cos x = 3/5, then you the adjacent side (A) is 3 and the hypothenuse (H) is 5.

2) The Pythagorean Theorem. Using the SOH CAH TOA abbreviations, the Pythagorean Theorem is A² + O² = H². So, for the first problem, you can get the opposite side (O) by 3² + O² = 5² , which gives O = 4. So sin x = 4/5.

We can do the same kind of thing for sin y = 24/25. So the Pythagorean Theorem is A² + 24² = 25², which gives A = 7. So, cos y = 7/25.

3) The sin, cos and tan angle addition and subtraction formulas. You can search on Google and find them. For example, sin(x-y) = sin x cos y - cos x sin y. For the first problem, sin (x-y) = 4/5 • 7/25 - 3/5 • 24/25 = 28/125 - 72/125 = -44/125. That's the answer.

You use the same kinds of techniques for the second problem.

If cot x = 6/5 then tan is the reciprocal, tan x = 5/6

If sec y = 3/2 then cos is the reciprocal, cos y = 2/3. To get the opposite side, 2² + O² = 3² , so O = √5. So tan y = √5/2

Looking up the tan addition formula, tan (x+y) = (tan x + tan y) / (1 - tan x tan y). So that's (5/6 + √5/2) / (1 - 5/6 • √5/2), which gives (5/6 + √5/2) / (1 - 5√5/12). This is the answer, but you could simplify it.

To simplify, a good technique is to make all the fractions have the same denominators, giving (10/12 + 6√5/12) / (12/12 - 5√5/12). Then you can get rid of all the 12s. And the simplified answer is (10 + 6√5) / (12 - 5√5)