Conditional Identities

Now that we know compound identities, multiple angles and submultiple angles identities, we can learn conditional identities.

Conditional identities in trigonometry are mathematical statements that hold true under specific conditions involving the variables within the identity. These conditions often dictate the values or relationships between the angles or other variables in the identity.

For example, let's consider the conditional identity \( (1 + \tan A)(1 + \tan B) = 2 \), under the condition \( A + B = \frac{\pi}{4} \). The identity is not true for all possible values of \(A\) and \(B\). It is true ony when \( A + B = \frac{\pi}{4} \).

Given \( A + B = \frac{\pi}{4} \), we know that \( \tan(A + B) = \tan\left(\frac{\pi}{4}\right) = 1 \). Thus,

\[ 1 = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

Rearranging this, we get:

\[ 1 - \tan A \tan B = \tan A + \tan B \]

\[ \tan A + \tan B + \tan A \tan B = 1 \]

Now, we add 1 to both sides of this equation:

\[ \tan A + \tan B + \tan A \tan B + 1 = 2 \]

This can be rewritten as:

\[ (1 + \tan A)(1 + \tan B) = 2 \]

Thus, under the condition \( A + B = \frac{\pi}{4} \), we demonstrate that \( (1 + \tan A)(1 + \tan B) = 2 \).

So, one can now claim that \((1+\tan17^\circ)(1+\tan28^\circ)=2\)

Conditional Identities in a triangle

In a triangle named ABC, the angles A, B, and C add up to 180 degrees. This fact gives us some important identities satisfied by the angles A, B, and C. It's good to remember these identities. Let's start by looking at these important identities and learn how to prove them. Most of these proofs are quite similar. As we go through them, try to spot the patterns that show up.

1. Given that in a triangle ABC, \( A + B + C = 180^\circ \), prove that \( \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \).

   Proof

   \[ \begin{align*} \sin 2A + \sin 2B + \sin 2C &= \sin 2A + 2 \sin (B+C) \cos (B-C)\\ &= \sin 2A + 2 \sin(180^\circ - A) \cos (B - C) \\ &= 2\sin A \cos A + 2 \sin A \cos (B - C) \\ &= 2\sin A (\cos A + \cos (B - C)) \\ &= 2\sin A (\cos (\pi - (B + C)) + \cos (B - C)) \\ &= 2\sin A (-\cos(B + C) + \cos (B - C)) \\ &= 2\sin A (2\sin B \sin C) \\ &= 4 \sin A \sin B \sin C. \end{align*} \]

   Step 1: Consider the Sum of Sines

   We start with the left-hand side of the desired identity:

   \[ \sin 2A + \sin 2B + \sin 2C \]

   Step 2: Use the Sine Addition Formula for \( \sin 2B \) and \( \sin 2C \)

   We can write \( \sin 2B \) and \( \sin 2C \) as \( 2 \sin(B + C) \cos(B - C) \). Therefore, we have:

   \[ \sin 2A + 2 \sin(B + C) \cos(B - C) \]

   The term \( 2 \sin(B + C) \) can be replaced by \( 2 \sin(180^\circ - A) \), which is equal to \( 2 \sin A \) since sine is a periodic function:

   \[ \sin 2A + 2 \sin A \cos(B - C) \]

   Step 3: Break Down \( \sin 2A \) into \( 2 \sin A \cos A \)

   Now, we use the double-angle identity for \( \sin 2A \):

   \[ 2 \sin A \cos A + 2 \sin A \cos(B - C) \]

   Step 4: Factor Out \( 2 \sin A \)

   Factoring out \( 2 \sin A \) from both terms, we get:

   \[ 2 \sin A (\cos A + \cos(B - C)) \]

   Step 5: Use the Cosine of Sum of Angles Identity

   We can express \( \cos A \) as \( \cos(180^\circ - (B + C)) \), which simplifies to \( -\cos(B + C) \) since cosine is negative in the second quadrant. Hence, we have:

   \[ 2 \sin A (-\cos(B + C) + \cos(B - C)) \]

   Step 7: Use the Sum to Product Identities

   The expression inside the parenthesis can be transformed using the sum-to-product identities:

   \[ 2 \sin A (2 \sin B \sin C) \]

   Step 8: Simplify and Reach the Conclusion

   Simplifying the expression, we finally arrive at:

   \[ 4 \sin A \sin B \sin C \]

   This completes the proof that \( \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \), given that \( A + B + C = 180^\circ \) in a triangle.

2. Given that in a triangle ABC, \( A + B + C = \pi \), prove that \( \cos 2A + \cos 2B + \cos 2C = -1 - 4 \cos A \cos B \cos C \).

   Proof:

   To prove the identity given \( A + B + C = \pi \):

   \[ \begin{align*} \cos 2A + \cos 2B + \cos 2C &= \cos 2A + (\cos 2B + \cos 2C) \\ &= \cos 2A + 2\cos(B + C) \cos(B - C) \\ &= \cos 2A + 2\cos(\pi - A) \cos(B - C) \\ &= \cos 2A - 2\cos A \cos(B - C) \\ &= -1 + 2\cos^2 A - 2\cos A \cos(B - C) \\ &= -1 + 2\cos A (\cos A - \cos(B - C)) \\ &= -1 + 2\cos A (\cos(\pi - (B + C)) - \cos(B - C)) \\ &= -1 - 2\cos A (\cos(B + C) + \cos(B - C)) \\ &= -1 - 2\cos A (2\cos B \cos C) \\ &= -1 - 4\cos A \cos B \cos C. \end{align*} \]

   Here's a step-by-step explanation:

   1. Expression Simplification: Start with the left-hand side of the identity, \(\cos 2A + \cos 2B + \cos 2C\).

   2. Apply Sum-to-Product Identities: Combine \(\cos 2B\) and \(\cos 2C\) using the sum-to-product identities to get \(2\cos(B + C) \cos(B - C)\).

   3. Substitute for \(B + C\): Replace \(B + C\) with \(\pi - A\), which simplifies to \(2\cos(\pi - A) \cos(B - C)\).

   4. Simplify Using Trigonometric Identities: Since \(\cos(\pi - A) = -\cos A\), the expression becomes \(\cos 2A - 2\cos A \cos(B - C)\).

   5. Break Down \(\cos 2A\): Decompose \(\cos 2A\) into \(-1 + 2\cos^2 A\) using the double-angle formula.

   6. Combine Like Terms: Collect the \(\cos A\) terms together, which gives \(2\cos A (\cos A - \cos(B - C))\).

   7. Use Angle Difference for \(\cos A\): Recognize that \(\cos A = \cos(\pi - (B + C))\) and apply this to the expression.

   8. Apply Sum-to-Product Identities Again: Rewrite the combined \(\cos\) terms as \(-2\cos A (\cos(B + C) + \cos(B - C))\).

   9. Final Simplification: Note that \(\cos(B + C) + \cos(B - C) = 2\cos B \cos C\), which leads to the final expression \(-1 - 4\cos A \cos B \cos C\).

   This proves the identity \(\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C\) for a triangle where \(A + B + C = \pi\).

3. Given that \( A + B + C = \pi \) in a triangle, prove that \( \sin A + \sin B + \sin C = 4 \cos\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right) \cos\left(\frac{C}{2}\right) \).

   Proof:

   To prove that \(\sin A + \sin B + \sin C = 4 \cos\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right) \cos\left(\frac{C}{2}\right)\) given \(A + B + C = \pi\):

   \[ \begin{align*} \sin A + \sin B + \sin C &= \sin A + 2 \sin\left(\frac{B + C}{2}\right) \cos\left(\frac{B - C}{2}\right) \\ &= \sin A + 2 \sin\left(\frac{\pi - A}{2}\right) \cos\left(\frac{B - C}{2}\right) \\ &= \sin A + 2 \cos\left(\frac{A}{2}\right) \cos\left(\frac{B - C}{2}\right) \\ &= 2 \cos\left(\frac{A}{2}\right) \left(\sin\left(\frac{A}{2}\right) + \cos\left(\frac{B - C}{2}\right)\right) \\ &= 2 \cos\left(\frac{A}{2}\right) \left(\sin\left(\frac{\pi - (B + C)}{2}\right) + \cos\left(\frac{B - C}{2}\right)\right) \\ &= 2 \cos\left(\frac{A}{2}\right) \left(\cos\left(\frac{B + C}{2}\right) + \cos\left(\frac{B - C}{2}\right)\right) \\ &= 2 \cos\left(\frac{A}{2}\right) \times 2 \cos\left(\frac{B}{2}\right) \cos\left(\frac{C}{2}\right) \\ &= 4 \cos\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right) \cos\left(\frac{C}{2}\right). \end{align*} \]

   This completes the proof of the given trigonometric identity.

4. Prove that in any triangle with angles \(A\), \(B\), and \(C\), where \( A + B + C = \pi \), the following identity holds true:

   \[ \cos A + \cos B + \cos C = 1 + 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \]
   Proof:

   Start with the left hand side:

   \[ \begin{align*} \cos A + \cos B + \cos C &= \cos A + 2 \cos\frac{B + C}{2} \cos\frac{B - C}{2} \\ &= \cos A + 2 \cos\frac{\pi - A}{2} \cos\frac{B - C}{2} \\ &= \cos A + 2 \sin\frac{A}{2} \cos\frac{B - C}{2} \\ &= 1 - 2\sin^2\frac{A}{2} + 2\sin\frac{A}{2} \cos\frac{B - C}{2} \\ &= 1 - 2\sin\frac{A}{2} (\sin\frac{A}{2} - \cos\frac{B - C}{2}) \\ &= 1 - 2\sin\frac{A}{2} (\sin\frac{A}{2} - \cos\frac{\pi - (B + C)}{2}) \\ &= 1 - 2\sin\frac{A}{2} (\cos\frac{B + C}{2} - \cos\frac{B - C}{2}) \\ &= 1 - 2\sin\frac{A}{2} \times (-2\sin\frac{B}{2} \sin\frac{C}{2}) \\ &= 1 + 4\sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} \end{align*} \]

5. Prove that in any triangle with angles \(A\), \(B\), and \(C\), where \( A + B + C = 180^\circ \), the following identity holds true:

   \[ \sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C. \]
   Proof:

   Begin from the left hand side:

   \[\begin{align*} \sin^2 A + \sin^2 B + \sin^2 C &= \frac{1}{2}(2\sin^2 A + 2\sin^2 B + 2\sin^2 C) \\ &= \frac{1}{2}(1 - \cos 2A + 1 - \cos 2B + 1 - \cos 2C) \\ &= \frac{1}{2}(3 - \cos 2A - \cos 2B - \cos 2C) \\ &= \frac{1}{2}(4 + 4(\cos A \cos B \cos C)) \\ &= 2 + 2\cos A \cos B \cos C \end{align*} \]

   Note that in step 4 we used: \( \cos 2A + \cos 2B + \cos 2C = -1 - 4 \cos A \cos B \cos C \)

6. Prove that in any triangle with angles \(A\), \(B\), and \(C\), where \( A + B + C = 180^\circ \), the following identity holds true:

   \[ \tan A + \tan B + \tan C = \tan A \tan B \tan C. \]
   Proof:

   We know that,

   \[\begin{align*} &A + B + C = \pi \\ &\Rightarrow A + B = \pi - C \\ &\Rightarrow \tan(A + B) = \tan(\pi - C) \\ &\Rightarrow \frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C \\ &\Rightarrow \tan A + \tan B = -\tan C + \tan A \tan B \tan C \\ &\Rightarrow \tan A + \tan B + \tan C = \tan A \tan B \tan C \end{align*}\]

7. Given a triangle with angles \(A\), \(B\), and \(C\) where \(A + B + C = \pi\):

   \[ \cot A \cot B + \cot B \cot C + \cot C \cot A = 1 \]
   Proof:

   Starting with the given identity:

   \[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]

   Divide both sides by \( \tan A \tan B \tan C \) to get:

   \[ \frac{\tan A}{\tan A \tan B \tan C} + \frac{\tan B}{\tan A \tan B \tan C} + \frac{\tan C}{\tan A \tan B \tan C} = \frac{\tan A \tan B \tan C}{\tan A \tan B \tan C} \]

   Simplify each term:

   \[ \frac{1}{\tan B \tan C} + \frac{1}{\tan A \tan C} + \frac{1}{\tan A \tan B} = 1 \]

   Recall the reciprocal identities \( \cot \theta = \frac{1}{\tan \theta} \) and apply them to each term:

   \[ \cot B \cot C + \cot A \cot C + \cot A \cot B = 1 \]

   This completes the proof that:

   \[ \cot A \cot B + \cot B \cot C + \cot C \cot A = 1 \]

8. Prove that in any triangle with angles \( A \), \( B \), and \( C \) where \( A + B + C = 180^\circ \), the following identity holds:

   \[ \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right) + \tan\left(\frac{B}{2}\right)\tan\left(\frac{C}{2}\right) + \tan\left(\frac{C}{2}\right)\tan\left(\frac{A}{2}\right) = 1. \]
   Proof:

   Starting with the sum of angles in a triangle:

   \[ A + B + C = \pi \]

   This leads to the equation:

   \[ \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}. \]

   Hence, we can write:

   \[ \frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}. \]

   Applying the tangent function to both sides gives us:

   \[ \tan\left(\frac{A}{2} + \frac{B}{2}\right) = \tan\left(\frac{\pi}{2} - \frac{C}{2}\right). \]

   Using the tangent addition formula, we have:

   \[ \frac{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right)}{1 - \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right)} = \cot\left(\frac{C}{2}\right) \]

   which simplifies to:

   \[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) = \cot\left(\frac{C}{2}\right) \left(1 - \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right)\right). \]

   Multiplying both sides by \( \tan\left(\frac{C}{2}\right) \) yields:

   \[ \tan\left(\frac{A}{2}\right)\tan\left(\frac{C}{2}\right) + \tan\left(\frac{B}{2}\right)\tan\left(\frac{C}{2}\right) = 1 - \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right). \]

   Rearranging terms, we get:

   \[ \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right) + \tan\left(\frac{B}{2}\right)\tan\left(\frac{C}{2}\right) + \tan\left(\frac{C}{2}\right)\tan\left(\frac{A}{2}\right) = 1. \]

   This concludes the proof.

9. Prove that for any triangle with angles \(A\), \(B\), and \(C\) where \(A + B + C = 180^\circ\), the following identity holds:

   \[ \cot\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right) + \cot\left(\frac{C}{2}\right) = \cot\left(\frac{A}{2}\right) \cot\left(\frac{B}{2}\right) \cot\left(\frac{C}{2}\right). \]
   Proof:

   Given the previous identity:

   \[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) + \tan\left(\frac{C}{2}\right) = \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right), \]

   we divide both sides by \( \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right) \):

   \[ \frac{1}{\tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right)} + \frac{1}{\tan\left(\frac{A}{2}\right) \tan\left(\frac{C}{2}\right)} + \frac{1}{\tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right)} = 1. \]

   By using the cotangent identity \( \cot(\theta) = \frac{1}{\tan(\theta)} \), we can rewrite the equation as:

   \[ \cot\left(\frac{A}{2}\right) \cot\left(\frac{B}{2}\right) + \cot\left(\frac{B}{2}\right) \cot\left(\frac{C}{2}\right) + \cot\left(\frac{C}{2}\right) \cot\left(\frac{A}{2}\right) = 1. \]

10. Prove that in any triangle with angles \(A\), \(B\), and \(C\) where \(A + B + C = \pi\), the following identity holds:

    \[ \cos\left(\frac{A}{2}\right) + \cos\left(\frac{B}{2}\right) + \cos\left(\frac{C}{2}\right) = 4\cos\left(\frac{\pi - A}{4}\right)\cos\left(\frac{\pi - B}{4}\right)\cos\left(\frac{\pi - C}{4}\right). \]
    Proof:

    Begin from the left hand side:

    \[ \begin{align*} \cos \frac{A}{2} + \cos \frac{B}{2} + \cos \frac{C}{2} &= \cos \frac{A}{2} + 2 \cos \frac{B+C}{4} \cos \frac{B-C}{4} \\ &= \cos \frac{A}{2} + 2 \cos \frac{\pi - A}{4} \cos \frac{B-C}{4} \\ &= \sin \frac{\pi - A}{2} + 2 \cos \frac{\pi - A}{4} \cos \frac{B-C}{4} \\ &= 2 \sin \frac{\pi - A}{4} \cos \frac{\pi - A}{4} + 2 \cos \frac{\pi - A}{4} \cos \frac{B-C}{4} \\ &= 2 \cos \frac{\pi - A}{4} \left( \sin \frac{\pi - A}{4} + \cos \frac{B-C}{4} \right) \\ &= 2 \cos \frac{\pi - A}{4} \left( \cos \left(\frac{\pi}{2} - \frac{\pi - A}{4}\right) + \cos \frac{B-C}{4} \right) \\ &= 2 \cos \frac{\pi - A}{4} \left( \cos \frac{\pi + A}{4} + \cos \frac{B-C}{4} \right) \\ &= 2 \cos \frac{\pi - A}{4} \left( \cos \frac{2\pi - (B+C)}{4} + \cos \frac{B-C}{4} \right) \\ &= 2 \cos \frac{\pi - A}{4} \left( \cos \frac{(\pi - B) + (\pi - C)}{4} + \cos \frac{(\pi - C) - (\pi - B)}{4} \right) \\ &= 2 \cos \frac{\pi - A}{4} \left( 2 \cos \frac{\pi - B}{4} \cos \frac{\pi - C}{4} \right) \\ &= 4 \cos \frac{\pi - A}{4} \cos \frac{\pi - B}{4} \cos \frac{\pi - C}{4}. \end{align*} \]

More complicated conditional identities

Here are some examples which are more complicated.

Example 1:

If \( A + B + C = 180^\circ \), then prove that:

\[ \frac{\tan{\frac{1}{2}A}}{\tan{\frac{1}{2}C}} = \frac{1 - \cos A + \cos B + \cos C}{1 - \cos C + \cos A + \cos B} \]

Solution

\[ \begin{align*} &=\frac{2\sin^2\left(\frac{A}{2}\right) + 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{B-C}{2}\right)}{2\sin^2\left(\frac{C}{2}\right) + 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)} \\ \\ &=\frac{2\sin\left(\frac{A}{2}\right)\left(\sin\left(\frac{A}{2}\right) + \cos\left(\frac{B-C}{2}\right)\right)}{2\sin\left(\frac{C}{2}\right)\left(\sin\left(\frac{C}{2}\right) + \cos\left(\frac{A-B}{2}\right)\right)} \\ \\ &=\frac{\sin\left(\frac{A}{2}\right)}{\sin\left(\frac{C}{2}\right)} \times \left(\frac{\cos\left(\frac{B+C}{2}\right) + \cos\left(\frac{B-C}{2}\right)}{\cos\left(\frac{A+B}{2}\right) + \cos\left(\frac{A-B}{2}\right)}\right) \\ \\ &=\frac{\sin\left(\frac{A}{2}\right)}{\sin\left(\frac{C}{2}\right)} \times \frac{2\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)}{2\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)} \\ \\ &=\frac{\tan\left(\frac{A}{2}\right)}{\tan\left(\frac{C}{2}\right)} \end{align*} \]

Example 2:

If \( A + B + C = \pi \), then prove that:

\[ \sum \sin^3 A = 3 \cos \frac{1}{2} A \cos \frac{1}{2} B \cos \frac{1}{2} C + \cos \frac{3}{2} A \cos \frac{3}{2} B \cos \frac{3}{2} C. \]

Solution:

Let us first prove that \(\sin3A+\sin3B+\sin3C = -4\cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2}\)

\[\begin{align*} \text{LHS} &= \sin(3A) + \sin(3B) + \sin(3C) \\ &= \sin(3A) + 2\sin\left(\frac{3B+3C}{2}\right)\cos\left(\frac{3B-3C}{2}\right) \\ &= \sin(3A) + 2\sin\left(\frac{3\pi - 3A}{2}\right)\cos\left(\frac{3B - 3C}{2}\right) \\ &= 2\sin(\frac{3A}{2})\cos(\frac{3A}{2}) - 2\cos\left(\frac{3A}{2}\right)\cos\left(\frac{3B - 3C}{2}\right) \\ &= 2\cos\left(\frac{3A}{2}\right)\left(\sin\left(\frac{3A}{2}\right) - \cos\left(\frac{3B-3C}{2}\right)\right) \\ &= 2\cos\left(\frac{3A}{2}\right)\left(\sin\left(\frac{3\pi - (3B+3C)}{2}\right) - \cos\left(\frac{3B-3C}{2}\right)\right) \\ &= 2\cos\left(\frac{3A}{2}\right)\left(-\cos\left(\frac{3B+3C}{2}\right) - \cos\left(\frac{3B-3C}{2}\right)\right) \\ &= -2\cos\left(\frac{3A}{2}\right)\left(2\cos\left(\frac{3B}{2}\right)\cos\left(\frac{3C}{2}\right)\right) \\ &= -4\cos\left(\frac{3A}{2}\right)\cos\left(\frac{3B}{2}\right)\cos\left(\frac{3C}{2}\right)\\ &= RHS\\ \end{align*}\]

To prove the given identity, we'll need to use the following formulas and trigonometric identities. Here's how you can proceed with the proof:

We know that:

\[ \sin^3 A = \frac{3\sin A - \sin 3A}{4} \]

\[ \sin^3 B = \frac{3\sin B - \sin 3B}{4} \]

\[ \sin^3 C = \frac{3\sin C - \sin 3C}{4} \]

We also know that for any angle \( A, B, C \) such that \( A + B + C = \pi \):

\[ \sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \]

\[ \sin 3A + \sin 3B + \sin 3C = -4\cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2} \]

We want to prove:

\[ \sum \sin^3 A = 3 \cos \frac{1}{2} A \cos \frac{1}{2} B \cos \frac{1}{2} C + \cos \frac{3}{2} A \cos \frac{3}{2} B \cos \frac{3}{2} C \]

Starting with the left-hand side, sum the expressions for \( \sin^3 A, \sin^3 B, \) and \( \sin^3 C \):

\[ \sin^3 A + \sin^3 B + \sin^3 C = \frac{3\sin A + 3\sin B + 3\sin C}{4} - \frac{\sin 3A + \sin 3B + \sin 3C}{4} \]

Using the given identities, replace \( \sin A + \sin B + \sin C \) and \( \sin 3A + \sin 3B + \sin 3C \) with their respective expressions:

\[ \sin^3 A + \sin^3 B + \sin^3 C = \frac{3(4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2})}{4} - \frac{-4\cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2}}{4} \]

Simplify the expression:

\[ \sin^3 A + \sin^3 B + \sin^3 C = 3\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} + \cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2} \]

Therefore, we have proved that:

\[ \sum \sin^3 A = 3 \cos \frac{1}{2} A \cos \frac{1}{2} B \cos \frac{1}{2} C + \cos \frac{3}{2} A \cos \frac{3}{2} B \cos \frac{3}{2} C \]

Example 3:

If \( A + B + C = 360^\circ \), and if

\[ \cos A = \frac{(d - a)(b - c)}{(d + a)(b + c)}, \quad \cos B = \frac{(d - b)(c - a)}{(d + b)(c + a)}, \quad \cos C = \frac{(d - c)(a - b)}{(d + c)(a + b)}, \]

then

\[ \tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} = \pm 1. \]

Solution:

Given:

\[ \cos A = \frac{(d - a)(b - c)}{(d + a)(b + c)}, \quad \cos B = \frac{(d - b)(c - a)}{(d + b)(c + a)}, \quad \cos C = \frac{(d - c)(a - b)}{(d + c)(a + b)} \]

Apply componendo and dividendo to each of the above equations:

For \( \cos A \), we get:

\[ \frac{1 + \cos A}{1 - \cos A} = \frac{(d + a)(b + c) + (d - a)(b - c)}{(d + a)(b + c) - (d - a)(b - c)} \]

For \( \cos B \), we get:

\[ \frac{1 + \cos B}{1 - \cos B} = \frac{(d + b)(c + a) + (d - b)(c - a)}{(d + b)(c + a) - (d - b)(c - a)} \]

For \( \cos C \), we get:

\[ \frac{1 + \cos C}{1 - \cos C} = \frac{(d + c)(a + b) + (d - c)(a - b)}{(d + c)(a + b) - (d - c)(a - b)} \]

Simplify the right hand side of each equation, and also knowing that \( \tan^2 \frac{A}{2} = \frac{1 - \cos A}{1 + \cos A} \), we can rewrite the equations above in terms of \( \tan^2 \frac{A}{2}, \tan^2 \frac{B}{2}, \) and \( \tan^2 \frac{C}{2} \):

For \( \cos A \), we have:

\[ \tan^2 \frac{A}{2} = \frac{dc+ab}{db+ac} \]

For \( \cos B \), we have:

\[ \tan^2 \frac{B}{2} = \frac{da+bc}{dc+ab} \]

For \( \cos C \), we have:

\[ \tan^2 \frac{C}{2} = \frac{db+ac}{da+bc} \]

Now, multiply all three equations:

\[ \tan^2 \frac{A}{2} \cdot \tan^2 \frac{B}{2} \cdot \tan^2 \frac{C}{2} = \left( \frac{dc+ab}{db+ac}\right) \left( \frac{da+bc}{dc+ab} \right) \left( \frac{db+ac}{da+bc}\right) = 1 \]

Simplify this expression to find \( \tan \frac{A}{2} \cdot \tan \frac{B}{2} \cdot \tan \frac{C}{2} \), and take the square root of both sides to get:

\[ \tan \frac{A}{2} \cdot \tan \frac{B}{2} \cdot \tan \frac{C}{2} = \pm 1 \]

According to the sum of angles identity for tangent, we have:

\[ \tan \left( \frac{A}{2} + \frac{B}{2} + \frac{C}{2} \right) = \frac{\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} - \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}{1 - (\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2})} \]

Given that \( A + B + C = 2\pi \), the angle \( \frac{A}{2} + \frac{B}{2} + \frac{C}{2} \) is equal to \( \pi \), and the tangent of \( \pi \) is 0. Therefore, the right-hand side of the identity is equal to 0. This implies that:

\[ \tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} - \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} = 0 \]

From the previous step, we already know that:

\[ \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} = \pm 1 \]

We can now substitute this into the equation above:

\[ \tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} = \pm 1 \]

This concludes the proof.5