Trigonometry Identities: Mixed Practice

Half of 60° is 30°; tan30° is √3/3.

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

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.

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

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

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

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

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

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

 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.

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

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

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

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

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

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

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

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