sin x = 1/4, x is in quadrant 2, find find sin x/2, cos x/2, tan x/2

To solve for \(\sin \frac{x}{2}\), \(\cos \frac{x}{2}\), and \(\tan \frac{x}{2}\) when \(\sin x = \frac{1}{4}\) and \(x\) is in quadrant 2, we'll use the **half-angle identities** for sine, cosine, and tangent:

### Step 1: Determine \(\cos x\)
We know that:
\[
\sin^2 x + \cos^2 x = 1
\]
Substitute \(\sin x = \frac{1}{4}\):
\[
\left(\frac{1}{4}\right)^2 + \cos^2 x = 1
\]
\[
\frac{1}{16} + \cos^2 x = 1
\]
\[
\cos^2 x = 1 - \frac{1}{16} = \frac{15}{16}
\]
\[
\cos x = \pm \frac{\sqrt{15}}{4}
\]
Since \(x\) is in the second quadrant, \(\cos x\) is negative:
\[
\cos x = -\frac{\sqrt{15}}{4}
\]

### Step 2: Use the half-angle identities

1. **Sine half-angle identity**:
\[
\sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}}
\]
Substitute \(\cos x = -\frac{\sqrt{15}}{4}\):
\[
\sin \frac{x}{2} = \pm \sqrt{\frac{1 - \left(-\frac{\sqrt{15}}{4}\right)}{2}} = \pm \sqrt{\frac{1 + \frac{\sqrt{15}}{4}}{2}}
\]
Simplify the expression:
\[
\sin \frac{x}{2} = \pm \sqrt{\frac{\frac{4 + \sqrt{15}}{4}}{2}} = \pm \sqrt{\frac{4 + \sqrt{15}}{8}}
\]
Since \(\frac{x}{2}\) is in the first quadrant (half of the second quadrant angle), \(\sin \frac{x}{2}\) is positive:
\[
\sin \frac{x}{2} = \sqrt{\frac{4 + \sqrt{15}}{8}}
\]

2. **Cosine half-angle identity**:
\[
\cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}}
\]
Substitute \(\cos x = -\frac{\sqrt{15}}{4}\):
\[
\cos \frac{x}{2} = \pm \sqrt{\frac{1 - \frac{\sqrt{15}}{4}}{2}} = \pm \sqrt{\frac{1 - \frac{\sqrt{15}}{4}}{2}}
\]
Simplify the expression:
\[
\cos \frac{x}{2} = \pm \sqrt{\frac{\frac{4 - \sqrt{15}}{4}}{2}} = \pm \sqrt{\frac{4 - \sqrt{15}}{8}}
\]
Since \(\frac{x}{2}\) is in the first quadrant, \(\cos \frac{x}{2}\) is positive:
\[
\cos \frac{x}{2} = \sqrt{\frac{4 - \sqrt{15}}{8}}
\]

3. **Tangent half-angle identity**:
\[
\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\sqrt{\frac{4 + \sqrt{15}}{8}}}{\sqrt{\frac{4 - \sqrt{15}}{8}}} = \sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}}
\]

Thus, the values are:
\[
\sin \frac{x}{2} = \sqrt{\frac{4 + \sqrt{15}}{8}}, \quad \cos \frac{x}{2} = \sqrt{\frac{4 - \sqrt{15}}{8}}, \quad \tan \frac{x}{2} = \sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}}
\]

Do you want details on simplifying these results or have further questions?

Here are 5 related questions for practice:
1. What is \(\sin \frac{x}{2}\) if \(\sin x = \frac{3}{5}\) and \(x\) is in quadrant 2?
2. If \(\cos x = -\frac{3}{5}\) in quadrant 2, find \(\sin \frac{x}{2}\).
3. How do you derive the half-angle identity for sine?
4. Can you use the double-angle identity to verify \(\cos x\) if \(\sin x = \frac{1}{4}\)?
5. What quadrant is \(\frac{x}{2}\) in if \(x\) is in the third quadrant?

**Tip:** Always check the quadrant to assign the correct sign when using trigonometric identities.

sin x = 1/4, x is in quadrant 2, find sin x/2, cos x/2, tan x/2

Learn how to solve the trigonometric problem of finding sin(x/2), cos(x/2), and tan(x/2) when sin x = 1/4 and x is in Quadrant 2. This problem involves half-angle identities, trigonometric properties, and Pythagorean identities, suitable for high school students in Grades 10-12.

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