Tangent Half Angle

The “one minus cosine over sine” version is the tangent half-angle.

Half of 210° is 105° (Quadrant II), where tangent is negative.

The three classic forms are sin/(1+cos), (1−cos)/sin, and the square-root ratio; cos/(1+sin) is not a standard half-angle tangent.

The “sine over one plus cosine” version is also the tangent half-angle.

With tan⁡θ=5/12 (so sin⁡=5/13, cos⁡=12/13), the half-angle tangent is (5/13)/(25/13)=1/5 

Half of 300° is 150°(Quadrant II), where tangent is −√3/3.

With tan⁡θ=3/4 (so sin⁡=3/5, cos⁡=4/5), the half-angle tangent is sin divided by 1+cos, which is (3/5)/(9/5)=1/3.

Half of 75° is 37.5°; the tangent there is about 0.767.

cosθ=−5/13 in Quadrant II gives sin⁡θ=12/13; then (12/13)/(1−5/13)=12/8=3/2, positive (half-angle in Quadrant I).

cosθ=12/13 in Quadrant IV gives sin⁡θ=−5/13; half-angle tangent is (−5/13)/(25/13)=−1/5

To find tan(θ/2) when sin(θ) = 24/25, we use the half-angle formula: tan(θ/2) = √((1 - cos(θ)) / (1 + cos(θ))). First, we determine cos(θ) using the Pythagorean identity: cos(θ) = √(1 - sin²(θ)) = √(1 - (24/25)²) = √(1 - 576/625) = √(49/625) = 7/25. Now substituting the values into the half-angle formula gives us tan(θ/2) = √((1 - 7/25) / (1 + 7/25)) = √((18/25) / (32/25)) = √(9/16) = 3/4. Therefore, we simplify and find that the correct answer is 7/24.

67.5°=135°/2; the exact tangent there is √2 + 1.

With cos⁡θ=0 in Quadrant I, you’re at 90°; half is 45° and the tangent is 1.

To find tan(θ/2) when tan(θ) = 1, we can use the half-angle formula: tan(θ/2) = (1 - cos(θ)) / sin(θ). Since tan(θ) = 1, we know that sin(θ) = cos(θ). Thus, θ = 45 degrees or π/4 radians, leading to tan(θ/2) = tan(22.5 degrees). We can also directly compute it using the identity tan(θ) = sin(θ) / cos(θ) and derive the necessary values. This results in tan(θ/2) = √2 - 1.

If tan⁡(θ/2)=2/3 (Quadrant I), then tan⁡θ=2(2/3)1−(2/3)2=12/5

sin⁡θ=24/25 and cos⁡θ=7/25 give (24/25)/(32/25)=3/4 

With 0<θ<π and cos⁡θ=0, you’re at 90°; half is 45°, and the tangent is 1.

If tanθ=1 in Quadrant I, then θ=45° and θ/2=22.5°; the exact tangent is √2 − 1.

With cos⁡θ=0.6 and sin⁡θ=0.8, the half-angle tangent is 0.8/(1.6)=0.5

To find tan(θ/2), we use the half-angle identity: tan(θ/2) = sin(θ) / (1 + cos(θ)). Since cos(θ) = 12/13, we can find sin(θ) using the Pythagorean identity: sin(θ) = -√(1 - cos²(θ)) = -√(1 - (12/13)²) = -5/13 (negative because θ is in quadrant IV). Then, substituting into the half-angle formula gives tan(θ/2) = (-5/13) / (1 + 12/13) = (-5/13) / (25/13) = -5/25 = -1/5. This corresponds to option D.