Solved UNIT 6 WORKSHEET 16 EVALUATING TRIG FUNCTIONS OF ANY ... - Free Printable

Let’s solve each part step by step. We’ll use the Pythagorean identity:
sin²θ + cos²θ = 1
And remember:
- tan θ = sin θ / cos θ
- cot θ = 1 / tan θ
- sec θ = 1 / cos θ
- csc θ = 1 / sin θ

Also, we must pay attention to the signs based on the quadrant (given by the signs of sin and cos).

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A) cos θ = 1/2, sin θ < 0

Since cos is positive and sin is negative → Quadrant IV

We know cos θ = adjacent/hypotenuse = 1/2 → so let’s imagine a right triangle with adjacent = 1, hypotenuse = 2.

Then opposite side = √(hyp² - adj²) = √(4 - 1) = √3

But since sin θ < 0 in QIV → sin θ = -√3 / 2

Now compute all six:

- sin θ = -√3 / 2
- cos θ = 1/2
- tan θ = sin/cos = (-√3/2) / (1/2) = -√3
- csc θ = 1/sin = 1 / (-√3/2) = -2/√3 = -2√3 / 3 (rationalized)
- sec θ = 1/cos = 1 / (1/2) = 2
- cot θ = 1/tan = 1 / (-√3) = -√3 / 3

✔ Check: All signs match QIV — sin-, cos+, tan- → correct.

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B) sin θ = -3/4, tan θ > 0

sin negative, tan positive → tan = sin/cos → if sin is neg and tan is pos, then cos must be negative → Quadrant III

Use identity: sin²θ + cos²θ = 1
→ (-3/4)² + cos²θ = 1
→ 9/16 + cos²θ = 1
→ cos²θ = 7/16
→ cos θ = -√7 / 4 (since QIII, cos negative)

Now:

- sin θ = -3/4
- cos θ = -√7 / 4
- tan θ = sin/cos = (-3/4) / (-√7/4) = 3/√7 = 3√7 / 7
- csc θ = 1/sin = 1 / (-3/4) = -4/3
- sec θ = 1/cos = 1 / (-√7/4) = -4/√7 = -4√7 / 7
- cot θ = 1/tan = √7 / 3

✔ Signs: sin-, cos-, tan+ → QIII → correct.

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C) sin θ = 1/4, cos θ > 0

sin positive, cos positive → Quadrant I

Use identity: sin²θ + cos²θ = 1
→ (1/4)² + cos²θ = 1
→ 1/16 + cos²θ = 1
→ cos²θ = 15/16
→ cos θ = √15 / 4 (positive in QI)

Now:

- sin θ = 1/4
- cos θ = √15 / 4
- tan θ = sin/cos = (1/4) / (√15/4) = 1/√15 = √15 / 15
- csc θ = 1/sin = 4
- sec θ = 1/cos = 4/√15 = 4√15 / 15
- cot θ = 1/tan = √15

✔ All positive → QI → correct.

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D) sec θ = 3, csc θ < 0

sec θ = 3 → cos θ = 1/3 (positive)
csc θ < 0 → sin θ < 0
So cos+, sin- → Quadrant IV

Use identity: sin²θ + cos²θ = 1
→ sin²θ + (1/3)² = 1
→ sin²θ + 1/9 = 1
→ sin²θ = 8/9
→ sin θ = -√8 / 3 = -2√2 / 3 (negative in QIV)

Now:

- sin θ = -2√2 / 3
- cos θ = 1/3
- tan θ = sin/cos = (-2√2/3) / (1/3) = -2√2
- csc θ = 1/sin = 1 / (-2√2/3) = -3/(2√2) = -3√2 / 4
- sec θ = 3 (given)
- cot θ = 1/tan = 1 / (-2√2) = -√2 / 4

✔ Signs: sin-, cos+, tan- → QIV → correct.

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E) tan θ = -√5 / 2, sin θ < 0

tan negative, sin negative → tan = sin/cos → if sin is neg and tan is neg, then cos must be positive → Quadrant IV

tan θ = opposite/adjacent = -√5 / 2 → but since we’re in QIV, let’s take opposite = -√5, adjacent = 2 (so tan = -√5/2)

Then hypotenuse = √(adj² + opp²) = √(4 + 5) = √9 = 3

So:

- sin θ = opp/hyp = -√5 / 3
- cos θ = adj/hyp = 2/3
- tan θ = -√5 / 2 (given)
- csc θ = 1/sin = -3/√5 = -3√5 / 5
- sec θ = 1/cos = 3/2
- cot θ = 1/tan = -2/√5 = -2√5 / 5

✔ Signs: sin-, cos+, tan- → QIV → correct.

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F) cos θ = √3 / 2, sin θ < 0

cos positive, sin negative → Quadrant IV

This is a standard angle! cos θ = √3/2 → reference angle 30° or π/6

In QIV, sin θ = -1/2

Check with identity:
sin²θ + cos²θ = (-1/2)² + (√3/2)² = 1/4 + 3/4 = 1 → good.

Now:

- sin θ = -1/2
- cos θ = √3 / 2
- tan θ = sin/cos = (-1/2) / (√3/2) = -1/√3 = -√3 / 3
- csc θ = 1/sin = -2
- sec θ = 1/cos = 2/√3 = 2√3 / 3
- cot θ = 1/tan = -√3

✔ Signs: sin-, cos+, tan- → QIV → correct.

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Final Answer:

A)
sin θ = -√3/2
cos θ = 1/2
tan θ = -√3
csc θ = -2√3/3
sec θ = 2
cot θ = -√3/3

B)
sin θ = -3/4
cos θ = -√7/4
tan θ = 3√7/7
csc θ = -4/3
sec θ = -4√7/7
cot θ = √7/3

C)
sin θ = 1/4
cos θ = √15/4
tan θ = √15/15
csc θ = 4
sec θ = 4√15/15
cot θ = √15

D)
sin θ = -2√2/3
cos θ = 1/3
tan θ = -2√2
csc θ = -3√2/4
sec θ = 3
cot θ = -√2/4

E)
sin θ = -√5/3
cos θ = 2/3
tan θ = -√5/2
csc θ = -3√5/5
sec θ = 3/2
cot θ = -2√5/5

F)
sin θ = -1/2
cos θ = √3/2
tan θ = -√3/3
csc θ = -2
sec θ = 2√3/3
cot θ = -√3