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lakkis [162]
3 years ago
5

If a and b are two angles in standard position in Quadrant I, find cos(a-b) for the given function values. sin a=3/5and cos b=12

/37
1) 153/185

2) 57/185

3) -57/185

4) -153/185

(Use the sum of identity for cosine)
Mathematics
1 answer:
goldfiish [28.3K]3 years ago
7 0

The identity in question is

cos(a - b) = cos(a) cos(b) + sin(a) sin(b)

so that

cos(a - b) = 12/37 cos(a) + 3/5 sin(b)

Since both a and b lie in the first quadrant, both cos(a) and sin(b) will be positive. Then it follows from the Pythagorean identity,

cos²(x) + sin²(x) = 1,

that

cos(a) = √(1 - sin²(a)) = 4/5

and

sin(b) = √(1 - cos²(b)) = 35/37

So,

cos(a - b) = 12/37 • 4/5 + 3/5 • 35/37 = 153/185

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Answer:

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Step-by-step explanation:

Given

g(x) = x^2 + 5

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Solving (a): (r o q)(2)

In function:

(r o g)(x) = r(g(x))

So, first we calculate g(2)

g(x) = x^2 + 5

g(2) = 2^2 + 5

g(2) = 4 + 5

g(2) = 9

Next, we calculate r(g(2))

Substitute 9 for g(2)in r(g(2))

r(q(2)) = r(9)

This gives:

r(x) = \sqrt{x + 7}

r(9) = \sqrt{9 +7{

r(9) = \sqrt{16}{

r(9) = 4

Hence:

(r o g)(2) = 4

Solving (b): (q o r)(2)

So, first we calculate r(2)

r(x) = \sqrt{x + 7}

r(2) = \sqrt{2 + 7}

r(2) = \sqrt{9}

r(2) = 3

Next, we calculate g(r(2))

Substitute 3 for r(2)in g(r(2))

g(r(2)) = g(3)

g(x) = x^2 + 5

g(3) = 3^2 + 5

g(3) = 9 + 5

g(3) = 14

Hence:

(q o r)(2) = 14

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Answer:

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