Solution of Triangles: Half Angle Formula

The solution of triangles is an important aspect of trigonometry that involves finding the unknown sides and angles of a triangle when certain sides and angles are known. The half angle formulas are particularly useful when dealing with oblique triangles, which are non-right triangles.

Half Angle Formulas

The half angle formulas are derived from the cosine and sine of double angles and are used to calculate the trigonometric functions of half angles. These formulas are particularly useful in solving triangles when we know the three sides and want to find the angles.

Sine Half Angle Formula

The sine of half an angle, $\sin\left(\frac{\alpha}{2}\right)$, can be found using the following formula:

$$ \sin\left(\frac{\alpha}{2}\right) = \pm\sqrt{\frac{1 - \cos(\alpha)}{2}} $$

Cosine Half Angle Formula

Similarly, the cosine of half an angle, $\cos\left(\frac{\alpha}{2}\right)$, can be expressed as:

$$ \cos\left(\frac{\alpha}{2}\right) = \pm\sqrt{\frac{1 + \cos(\alpha)}{2}} $$

Tangent Half Angle Formula

The tangent of half an angle, $\tan\left(\frac{\alpha}{2}\right)$, can be found using the formula:

$$ \tan\left(\frac{\alpha}{2}\right) = \frac{1 - \cos(\alpha)}{\sin(\alpha)} = \frac{\sin(\alpha)}{1 + \cos(\alpha)} $$

Application in Triangles

When solving for the angles of a triangle with known sides a, b, and c, we can use the Law of Cosines to find one of the angles and then apply the half angle formulas to find the other angles.

Law of Cosines

For a triangle with sides a, b, and c, and opposite angles A, B, and C respectively, the Law of Cosines states:

$$ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} $$

Once we have $\cos(A)$, we can use the half angle formulas to find $\sin\left(\frac{A}{2}\right)$ and $\cos\left(\frac{A}{2}\right)$.

Differences and Important Points

Aspect	Description
Usage	Half angle formulas are used when we need to find the trigonometric functions of half an angle.
Sign Ambiguity	The formulas yield both positive and negative results. The correct sign depends on the quadrant.
Relationship with Sides	In a triangle, half angle formulas can be used after applying the Law of Cosines or Sines.
Application in Calculations	Useful in calculus for integration and solving trigonometric integrals.

Examples

Example 1: Using Sine Half Angle Formula

Given a triangle with sides a = 7, b = 24, and c = 25, find $\sin\left(\frac{A}{2}\right)$.

First, use the Law of Cosines to find $\cos(A)$:

$$ \cos(A) = \frac{24^2 + 25^2 - 7^2}{2 \cdot 24 \cdot 25} = \frac{576 + 625 - 49}{1200} = \frac{1152}{1200} = \frac{96}{100} = 0.96 $$

Now, apply the sine half angle formula:

$$ \sin\left(\frac{A}{2}\right) = \pm\sqrt{\frac{1 - \cos(A)}{2}} = \pm\sqrt{\frac{1 - 0.96}{2}} = \pm\sqrt{\frac{0.04}{2}} = \pm\sqrt{0.02} $$

Since $\frac{A}{2}$ is an acute angle, we take the positive value:

$$ \sin\left(\frac{A}{2}\right) = \sqrt{0.02} \approx 0.1414 $$

Example 2: Using Cosine Half Angle Formula

Given the same triangle, find $\cos\left(\frac{A}{2}\right)$.

Using the cosine half angle formula:

$$ \cos\left(\frac{A}{2}\right) = \pm\sqrt{\frac{1 + \cos(A)}{2}} = \pm\sqrt{\frac{1 + 0.96}{2}} = \pm\sqrt{\frac{1.96}{2}} = \pm\sqrt{0.98} $$

Again, since $\frac{A}{2}$ is an acute angle, we take the positive value:

$$ \cos\left(\frac{A}{2}\right) = \sqrt{0.98} \approx 0.9899 $$

These examples illustrate how half angle formulas can be applied to solve for the angles of a triangle when the sides are known. It is important to remember the sign ambiguity and to choose the correct sign based on the context of the problem.