Trigonometry Identities: Mixed Practice

1) Evaluate tan(22.5°) exactly.

√2 − 1

√2 + 1

1/√2

1/2

2) If tan(θ/2) = 2/3, find sinθ.

5/13

12/13

3/5

4/5

To find sinθ from tan(θ/2) = 2/3, we use the double angle identity for sine: sinθ = 2sin(θ/2)cos(θ/2). First, we find sin(θ/2) and cos(θ/2) using the tangent. If tan(θ/2) = 2/3, we can think of this as a right triangle where the opposite side is 2 and the adjacent side is 3. By using the Pythagorean theorem, the hypotenuse would be √(2² + 3²) = √13. Therefore, sin(θ/2) = 2/√13 and cos(θ/2) = 3/√13. Then we calculate sinθ = 2 * (2/√13) * (3/√13) = 12/13.

Explanation

To find sinθ from tan(θ/2) = 2/3, we use the double angle identity for sine: sinθ = 2sin(θ/2)cos(θ/2). First, we find sin(θ/2) and cos(θ/2) using the tangent. If tan(θ/2) = 2/3, we can think of this as a right triangle where the opposite side is 2 and the adjacent side is 3. By using the Pythagorean theorem, the hypotenuse would be √(2² + 3²) = √13. Therefore, sin(θ/2) = 2/√13 and cos(θ/2) = 3/√13. Then we calculate sinθ = 2 * (2/√13) * (3/√13) = 12/13.

3) If θ = 300°, determine the sign of sin(θ/2).

Positive

Negative

Zero

Undefined

Half of 300° is 150° (Quadrant II), where sine is positive.

Explanation

Half of 300° is 150° (Quadrant II), where sine is positive.

4) If θ = 60°, compute tan(θ/2).

√3/3

√3

1

0

Half of 60° is 30°; tan30° is √3/3.

Explanation

Half of 60° is 30°; tan30° is √3/3.

5) Which formula expresses tan(θ/2)?

Sinθ/(1+cosθ)

(1+cosθ)/sinθ

Cosθ/(1+sinθ)

Cotθ

A standard form for tangent of the half-angle is “sine over one plus cosine.”

Explanation

A standard form for tangent of the half-angle is “sine over one plus cosine.”

6) Evaluate sin(22.5°) exactly.

√(2+√2)/2

√(2−√2)/2

√2/2

(√6−√2)/4

The exact value of sin⁡22.5° is √(2−√2)/2.

Explanation

The exact value of sin⁡22.5° is √(2−√2)/2.

7) Simplify sin²(θ/2) in terms of θ.

(1+cosθ)/2

(1−cosθ)/2

(1+sinθ)/2

(1−sinθ)/2

The square of the half-angle sine is the “one minus cosine” form.

Explanation

The square of the half-angle sine is the “one minus cosine” form.

8) If cosθ = 0 and 0 < θ < π, compute cos(θ/2).

0

√2/2

1

Undefined

With 0<θ<π And cos⁡θ=0, you’re at 90°; half is 45°, so cosine is √2/2.

Explanation

With 0<θ<π And cos⁡θ=0, you’re at 90°; half is 45°, so cosine is √2/2.

9) Which formula expresses sin(θ/2)?

±√((1+cosθ)/2)

±√((1−cosθ)/2)

±√((1+sinθ)/2)

±√((1−sinθ)/2)

The sine half-angle uses the “one minus cosine” form with a ± sign for the quadrant.

Explanation

The sine half-angle uses the “one minus cosine” form with a ± sign for the quadrant.

10) If cosθ = 4/5 and θ is in quadrant I, find sin(θ/2).

2/√5

3/5

1/√10

√5/5

With cos⁡θ=4/5 in Quadrant I, the half-angle sine is positive and equals 1/√10.

Explanation

With cos⁡θ=4/5 in Quadrant I, the half-angle sine is positive and equals 1/√10.

11) If tan(θ/2) = 2/3, find cosθ.

5/13

12/13

3/5

−5/13

To find cos(theta) using tan(theta/2), we employ the half-angle identity: cos(theta) = 1 - tan^2(theta/2) / (1 + tan^2(theta/2)). Here, tan(theta/2) = 2/3, so tan^2(theta/2) = (2/3)^2 = 4/9. Substituting into the formula gives cos(theta) = (1 - 4/9) / (1 + 4/9) = (5/9) / (13/9) = 5/13.

Explanation

To find cos(theta) using tan(theta/2), we employ the half-angle identity: cos(theta) = 1 - tan^2(theta/2) / (1 + tan^2(theta/2)). Here, tan(theta/2) = 2/3, so tan^2(theta/2) = (2/3)^2 = 4/9. Substituting into the formula gives cos(theta) = (1 - 4/9) / (1 + 4/9) = (5/9) / (13/9) = 5/13.

12) If sinθ = 24/25 (quadrant I), find cos(θ/2).

2√2/5

3/5

4/5

√2/5

With sin⁡θ=24/25 (Quadrant I) and cos⁡θ=7/25, the half-angle cosine is 4/5.

Explanation

With sin⁡θ=24/25 (Quadrant I) and cos⁡θ=7/25, the half-angle cosine is 4/5.

13) Simplify cos²(θ/2) in terms of θ.

(1−sinθ)/2

(1−cosθ)/2

(1+sinθ)/2

(1+cosθ)/2

The square of the half-angle cosine is the “one plus cosine” form.

Explanation

The square of the half-angle cosine is the “one plus cosine” form.

14) If θ is in quadrant II and cosθ = −12/13, find sin(θ/2).

5/√26

−5/√26

12/√26

−12/√26

In Quadrant II with cos⁡θ=−12/13, the half-angle sine is positive 5/√26.

Explanation

In Quadrant II with cos⁡θ=−12/13, the half-angle sine is positive 5/√26.

15) If 0 < θ < π and cosθ = 0.6, find sin(θ/2) in simplest radical form.

√(1/5)

√(2/5)

1/√5

2/√5

With 0<θ<π and cos⁡θ=0.6 (Quadrant I), the half-angle sine simplifies to 1/√5

Explanation

With 0<θ<π and cos⁡θ=0.6 (Quadrant I), the half-angle sine simplifies to 1/√5

16) Which formula expresses cos(θ/2)?

±√((1−sinθ)/2)

±√((1−cosθ)/2)

±√((1+sinθ)/2)

±√((1+cosθ)/2)

The cosine half-angle uses the “one plus cosine” form with a ± sign for the quadrant.

Explanation

The cosine half-angle uses the “one plus cosine” form with a ± sign for the quadrant.

17) If cosθ = 3/5 and θ is in quadrant IV, find cos(θ/2).

2/√5

−2/√5

1/√5

−1/√5

With cos⁡θ=3/5 in Quadrant IV, the half-angle cosine is negative with size 2/√5.

Explanation

With cos⁡θ=3/5 in Quadrant IV, the half-angle cosine is negative with size 2/√5.

18) Evaluate cos(22.5°) exactly.

√(2+√2)/2

√(2−√2)/2

√2/2

(√6−√2)/4

19) If sin(θ/2) = 3/5 and θ/2 is in quadrant I, find cosθ.

−16/25

−7/25

16/25

7/25

If sin⁡(θ/2)=3/5 in Quadrant I, then cos⁡θ=1−2(3/5)2=7/25

Explanation

If sin⁡(θ/2)=3/5 in Quadrant I, then cos⁡θ=1−2(3/5)2=7/25

20) If cos(θ/2) = 3/5 and θ/2 is in quadrant I, find cosθ.

7/25

−7/25

16/25

−16/25

If cos⁡(θ/2)=3/5 in Quadrant I, then cos⁡θ=2(3/5)2−1=−7/25

Explanation

If cos⁡(θ/2)=3/5 in Quadrant I, then cos⁡θ=2(3/5)2−1=−7/25