Compound Angles

Compound angles involve expressions where two or more angles are combined through addition or subtraction, such as \( \alpha + \beta \) or \( \alpha - \beta \). The trigonometric functions of these compound angles can be expressed in terms of the trigonometric functions of the individual angles. This is useful in various mathematical applications, as it allows the simplification of trigonometric expressions and the solving of equations that contain angles summed or subtracted from one another. The formulas for sine, cosine, and tangent of compound angles enable these conversions and are fundamental in both theoretical and applied mathematics.

Value of Trigonometric Functions for the sum of two angles

For sine (\(\sin\)):

\[ \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) \]

For cosine (\(\cos\)):

\[ \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) \]

For tangent (\(\tan\)):

\[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} \]

These formulas express the trigonometric functions of a sum of two angles in terms of the trigonometric functions of each individual angle.

Proof:

To prove we begin with proving cosine of sum. There are so many proofs available. In my opinion, this is probably the best one as it does not assume angles \(\alpha\) and \(\beta\) to be acute.

In the unit circle representation, let's define the points associated with the angles \(\alpha\), \(\beta\), and \(\alpha + \beta\):

Point \(P\) has coordinates \((\cos\alpha, \sin\alpha)\) corresponding to angle \(\alpha\).
Point \(Q\) represents the position of point \(P\) rotated further by angle \(\beta\), with coordinates \((\cos(\alpha + \beta), \sin(\alpha + \beta))\).
Point \(R\) has coordinates \((\cos(-\beta), \sin(-\beta))\) which can also be written as \((\cos\beta, -\sin\beta)\) due to the even-odd properties of the cosine and sine functions, corresponding to angle \(-\beta\).

Since the arc \(AQ\) and arc \(PR\) both subtend angle \(\beta\) at the center \(O\), we can infer that the chord lengths \(AQ\) and \(PR\) are equal, which leads to the equation:

\[ AQ^2 = PR^2 \]

Now, considering the distance formula:

\( AQ^2 = (\cos(\alpha + \beta) - 1)^2 + \sin^2(\alpha + \beta) \)

\( PR^2 = (\cos \alpha - \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 \)

Equating the two expressions:

\( (\cos(\alpha + \beta) - 1)^2 + \sin^2(\alpha + \beta) \)\( (\cos \alpha - \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 \)

Expanding both sides gives us:

\( \cos^2(\alpha + \beta) - 2\cos(\alpha + \beta) + 1 + \sin^2(\alpha + \beta) \)\(= \cos^2 \alpha - 2\cos \alpha \cos \beta + \cos^2 \beta + \sin^2 \alpha + 2\sin \alpha \sin \beta + \sin^2 \beta \)

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we can simplify both sides:

\( 1 - 2\cos(\alpha + \beta) + 1 = 1 - 2\cos \alpha \cos \beta + 1 + 2\sin \alpha \sin \beta \)

This simplifies to:

\( -2\cos(\alpha + \beta) = -2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta \)

Dividing through by -2:

\( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)

And that gives us the required formula for the cosine of a sum of two angles.

Using this formula we can derive the sine formula.

Start with the expression for \(\sin(\alpha + \beta)\) and use the co-function identity:

\(sin(\alpha + \beta) = \cos\left(\frac{\pi}{2} - (\alpha + \beta)\right)\)
Expand the inside of the cosine using the sum formula for cosine:

\(\cos\left(\frac{\pi}{2} - \alpha - \beta\right) = \cos\left(\frac{\pi}{2} - \alpha\right)\cos(-\beta) + \sin\left(\frac{\pi}{2} - \alpha\right)\sin(-\beta)\)
Apply the co-function identity and the fact that (\(\cos(-\beta) = \cos(\beta)\)) and (\(\sin(-\beta) = -\sin(\beta)\)):

\(\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)\)
Rearrange to match the standard form of the sine sum formula:

\(\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\)

Thus, we have derived the sine sum formula.

Now we derive the tangent of sum: To derive the formula for \(\tan(\alpha + \beta)\), we start with the sum formulas for sine and cosine and use the identity \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\):

Use the sine and cosine sum formulas: \(\tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}\)

\(\implies \tan(\alpha + \beta) = \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)}\)
Divide the numerator and the denominator by \(\cos(\alpha)\cos(\beta)\):

\(\tan(\alpha + \beta) = \frac{\frac{\sin(\alpha)}{\cos(\alpha)} + \frac{\sin(\beta)}{\cos(\beta)}}{1 - \frac{\sin(\alpha)}{\cos(\alpha)}\frac{\sin(\beta)}{\cos(\beta)}}\)

\(\implies \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\)

So, the formula for \(\tan(\alpha + \beta)\) is:

\(\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\)

This completes the proof. \(\blacksquare\)

Value of Trigonometric Functions for the difference of angles

For sine (\(\sin\)):

\[ \sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta \]

For cosine (\(\cos\)):

\[ \cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta \]

For tangent (\(\tan\)):

\[ \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} \]

Proof:

To derive the formulas for the difference of two angles for sine, cosine, and tangent functions, we can utilize the sum formulas for trigonometric functions and simply replace \(\beta\) with \(-\beta\). Here's how it works:

For Sine:

The sum formula for sine is:

\( \sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta \)

Replacing \(\beta\) with \(-\beta\), we get:

\( \sin(\alpha - \beta) = \sin\alpha \cos(-\beta) + \cos\alpha \sin(-\beta) \)

Since \(\cos(-\beta) = \cos\beta\) (because cosine is an even function) and \(\sin(-\beta) = -\sin\beta\)

\( \sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta \)

For Cosine:

The sum formula for cosine is:

\( \cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta \)

Again, replace \(\beta\) with \(-\beta\):

\( \cos(\alpha - \beta) = \cos\alpha \cos(-\beta) - \sin\alpha \sin(-\beta) \)

\( \implies \cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta \)

For Tangent:

The sum formula for tangent is:

\( \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} \)

Substituting \(\beta\) with \(-\beta\):

\( \tan(\alpha - \beta) = \frac{\tan\alpha + \tan(-\beta)}{1 - \tan\alpha \tan(-\beta)} \)

And since \(\tan(-\beta) = -\tan\beta\):

\( \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} \)

These are the formulas for the difference of two angles in trigonometry. They are directly derived from the sum formulas with the substitution of \(\beta\) with \(-\beta\).

Product-to-Sum identities:

The product of sine and cosine functions can be expressed as the sum of sine or cosine functions. The identities are as follows:

For the product of a sine and a cosine function of two angles:

\[ \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)] \]

\[ \cos A \sin B = \frac{1}{2}[\sin(A + B) - \sin(A - B)] \]
For the product of two cosine functions:

\[ \cos A \cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)] \]
For the product of two sine functions:

\[ \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \]

Proof:

To derive the identity for \(\sin A \cos B\), we start with the sum and difference formulas for sine:

\( \sin(A + B) = \sin A \cos B + \cos A \sin B \)

\( \sin(A - B) = \sin A \cos B - \cos A \sin B \)

Adding these two equations, we get:

\( \sin(A + B) + \sin(A - B) = 2 \sin A \cos B \)

Dividing by 2:

\( \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)] \)
For \(\cos A \sin B\), we subtract the second equation from the first:

\( \sin(A + B) - \sin(A - B) = 2 \cos A \sin B \)

Dividing by 2:

\( \cos A \sin B = \frac{1}{2}[\sin(A + B) - \sin(A - B)] \)
For \(\cos A \cos B\), we use the cosine sum and difference formulas:

\( \cos(A + B) = \cos A \cos B - \sin A \sin B \)

\( \cos(A - B) = \cos A \cos B + \sin A \sin B \)

Adding these two equations, we get:

\( \cos(A + B) + \cos(A - B) = 2 \cos A \cos B \)

Dividing by 2:

\( \cos A \cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)] \)
Lastly, for \(\sin A \sin B\), we subtract the first cosine equation from the second:

\( \cos(A - B) - \cos(A + B) = 2 \sin A \sin B \)

Dividing by 2:

\( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \)

This concludes the proof of the product-to-sum identities for sine and cosine functions. \(\blacksquare\)

A surprising use of these formulae: Prosthaphaeresis

Suppose you want to calculate \( 0.31425 \times 0.64256 \). We can achieve this using the table of natural cosines and the product-to-sum formulas for trigonometric functions, follow these steps:

Find the angles \( A \) and \( B \) for which \( \cos(A) = 0.31425 \) and \( \cos(B) = 0.64256 \) using the cosine table.
Use the product-to-sum formula: \( \cos(A) \cos(B) = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \)
Find the values of \( \cos(A + B) \) and \( \cos(A - B) \) from the table or calculate them using trigonometric identities if the exact values are not in the table.
Finally, calculate the product \( 0.31425 \times 0.64256 \) using the angles found.

Of course, this is an old way of multiplying two numbers. You can read about it here: Prosthaphaeresis

For us, the main purpose of these formulas is in simplifying expressions. Converting a product of sines and cosines to sum is sometimes desired during simplification of expressions or solving equations.

Sum to Product Formulas

Given two angles \( C \) and \( D \), the sum to product formulas for the trigonometric functions sine and cosine are as follows:

For the sum of two sine functions:

\[ \sin C + \sin D = 2 \sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right) \]
For the difference of two sine functions:

\[ \sin C - \sin D = 2 \cos\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right) \]
For the sum of two cosine functions:

\[ \cos C + \cos D = 2 \cos\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right) \]
For the difference of two cosine functions:

\[ \cos C - \cos D = -2 \sin\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right) \]

Proof:

To derive these formulas, we use the identities for the cosine of a sum and cosine of a difference, along with the corresponding identities for sine.

For the sine sum formula, we start with the sum and difference formulas for sine:

\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]

Adding these two equations, we get:

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

Now, let \( A = \frac{C + D}{2} \) and \( B = \frac{C - D}{2} \), then:

\[ \sin C + \sin D = 2 \sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right) \]

Similarly, by subtracting the sine difference formula from the sine sum formula, we can derive the second formula. The third and fourth formulas follow from the sum and difference formulas for cosine in an analogous manner.

Additional Conversion Formulae

I. For any angles \( A \) and \( B \), the following identities hold:

\( \cot(A + B) = \frac{\cot A \cdot \cot B - 1}{\cot A + \cot B} \)
\( \cot(A - B) = \frac{\cot A \cdot \cot B + 1}{\cot B - \cot A} \)

Proof

To prove the first identity, we start by expressing the cotangent in terms of the sine and cosine functions:

\[ \cot(A + B) = \frac{\cos(A + B)}{\sin(A + B)} \]

Using the angle sum formulas for sine and cosine, we have:

\[ \cos(A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B \]

\[ \sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B \]

Substituting these into our expression for \( \cot(A + B) \), we get:

\[ \cot(A + B) = \frac{\cos A \cdot \cos B - \sin A \cdot \sin B}{\sin A \cdot \cos B + \cos A \cdot \sin B} \]

Now, dividing both numerator and denominator by \( \cos A \cdot \cos B \), we get:

\[ \cot(A + B) = \frac{1 - \tan A \cdot \tan B}{\tan A + \tan B} \]

Finally, taking the reciprocal of both \( \tan A \) and \( \tan B \) (since \( \cot \theta = \frac{1}{\tan \theta} \)), we arrive at:

\[ \cot(A + B) = \frac{\cot A \cdot \cot B - 1}{\cot A + \cot B} \]

This proves the first identity.

For the second identity, we use a similar approach:

\[ \cot(A - B) = \frac{\cos(A - B)}{\sin(A - B)} \]

Using the angle difference formulas for sine and cosine, we have:

\[ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \]

\[ \sin(A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B \]

Substituting these into our expression for \( \cot(A - B) \), we get:

\[ \cot(A - B) = \frac{\cos A \cdot \cos B + \sin A \cdot \sin B}{\sin A \cdot \cos B - \cos A \cdot \sin B} \]

Dividing both numerator and denominator by \( \cos A \cdot \cos B \), we obtain:

\[ \cot(A - B) = \frac{1 + \tan A \cdot \tan B}{\tan B - \tan A} \]

Taking the reciprocal of both \( \tan A \) and \( \tan B \) gives us:

\[ \cot(A - B) = \frac{\cot A \cdot \cot B + 1}{\cot B - \cot A} \]

This proves the second identity.

\( \blacksquare \)

II. For any angles \( A \) and \( B \), the following trigonometric identities hold:

\( \sin(A + B) \cdot \sin(A - B) = \sin^2 A - \sin^2 B \)
\( \cos(A + B) \cdot \cos(A - B) = \cos^2 A - \sin^2 B \)
\( \tan(A + B) \cdot \tan(A - B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \cdot \tan^2 B} \)

Proof:

For identity 1:

We start by applying the sine addition and subtraction formulas:

\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]

Multiplying these two expressions together yields:

\[ \sin(A + B) \sin(A - B) = (\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B) \]

This expands to:

\[ = \sin^2 A \cos^2 B - \sin A \cos A \sin B \cos B + \sin A \cos A \sin B \cos B - \cos^2 A \sin^2 B \]

Simplifying, we get:

\[ = \sin^2 A \cos^2 B - \cos^2 A \sin^2 B \]

Since \( \cos^2 B = 1 - \sin^2 B \) and \( \cos^2 A = 1 - \sin^2 A \), we substitute to find:

\[ = \sin^2 A (1 - \sin^2 B) - (1 - \sin^2 A) \sin^2 B \]

\[ = \sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B \]

\[ = \sin^2 A - \sin^2 B \]

For identity 2:

Applying the cosine addition and subtraction formulas:

\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]

\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]

Multiplying these two expressions:

\[ \cos(A + B) \cos(A - B) = (\cos A \cos B - \sin A \sin B)(\cos A \cos B + \sin A \sin B) \]

Expands to:

\[ = \cos^2 A \cos^2 B - \sin^2 A \sin^2 B \]

Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:

\[ \cos^2 B = 1 - \sin^2 B \]

\[ \cos^2 A = 1 - \sin^2 A \]

Substituting these into our equation:

\[ = \cos^2 A (1 - \sin^2 B) - (1 - \cos^2 B) \sin^2 A \]

\[ = \cos^2 A - \cos^2 A \sin^2 B - \sin^2 A + \sin^2 A \cos^2 B \]

\[ = \cos^2 A - \sin^2 B \]

For identity 3:

We use the tangent addition and subtraction formulas:

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

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

Multiplying these two expressions:

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

This simplifies to:

\[ = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B} \]

Thus, we have proved all three identities.

\( \blacksquare \)

III. For any angle \( \theta \), the following trigonometric identities hold:

\( \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan \theta}{1 - \tan \theta} \)
\( \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} \)
\( \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{\pi}{4} - \theta\right) = 1 \)

Proof:

For identity 1:

The tangent addition formula states that for any angles \( x \) and \( y \):

\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]

Let \( x = \frac{\pi}{4} \) and \(y=\theta\), then we have:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan \theta}{1 - \tan\left(\frac{\pi}{4}\right) \tan \theta} \]

Knowing that \( \tan\left(\frac{\pi}{4}\right) = 1 \), we substitute this into the equation:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan \theta}{1 - 1 \cdot \tan \theta} \]

\[ = \frac{1 + \tan \theta}{1 - \tan \theta} \]

For identity 2:

Applying the tangent subtraction formula similarly:

\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan \theta}{1 + \tan\left(\frac{\pi}{4}\right) \tan \theta} \]

Again, \( \tan\left(\frac{\pi}{4}\right) = 1 \), so we have:

\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + 1 \cdot \tan \theta} \]

\[ = \frac{1 - \tan \theta}{1 + \tan \theta} \]

For identity 3:

Multiplying the results of identity 1 and identity 2 gives us:

\[ \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 + \tan \theta}{1 - \tan \theta} \cdot \frac{1 - \tan \theta}{1 + \tan \theta} \]

This simplifies to:

\( \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{\pi}{4} - \theta\right) = \frac{(1 + \tan \theta)(1 - \tan \theta)}{(1 - \tan \theta)(1 + \tan \theta)} \) \(= 1 \)

Thus, we have proved all three identities.

\( \blacksquare \)

IV. For any angle \( \theta \), the following trigonometric identities hold:

\( \sin \theta + \sin(\theta - 120^\circ) + \sin(\theta + 120^\circ) = 0 \)
\( \cos \theta + \cos(\theta - 120^\circ) + \cos(\theta + 120^\circ) = 0 \)

Proof:

For identity 1:

We apply the sine addition and subtraction formulas:

\[ \sin(\theta \pm 120^\circ) = \sin \theta \cos 120^\circ \pm \cos \theta \sin 120^\circ \]

Knowing that \( \cos 120^\circ = -\frac{1}{2} \) and \( \sin 120^\circ = \frac{\sqrt{3}}{2} \), we substitute these values in:

\[ \sin(\theta - 120^\circ) = \sin \theta \left(-\frac{1}{2}\right) - \cos \theta \left(\frac{\sqrt{3}}{2}\right) \]

\[ \sin(\theta + 120^\circ) = \sin \theta \left(-\frac{1}{2}\right) + \cos \theta \left(\frac{\sqrt{3}}{2}\right) \]

Adding the three sine terms:

\[ \sin \theta + \sin(\theta - 120^\circ) + \sin(\theta + 120^\circ) \]

\[ = \sin \theta - \frac{1}{2}\sin \theta - \frac{\sqrt{3}}{2}\cos \theta - \frac{1}{2}\sin \theta + \frac{\sqrt{3}}{2}\cos \theta \]

\[ = \sin \theta - \sin \theta \]

\[ = 0 \]

For identity 2:

We apply the cosine addition and subtraction formulas:

\[ \cos(\theta \pm 120^\circ) = \cos \theta \cos 120^\circ \mp \sin \theta \sin 120^\circ \]

Substitute the known values for \( \cos 120^\circ \) and \( \sin 120^\circ \):

\[ \cos(\theta - 120^\circ) = \cos \theta \left(-\frac{1}{2}\right) + \sin \theta \left(\frac{\sqrt{3}}{2}\right) \]

\[ \cos(\theta + 120^\circ) = \cos \theta \left(-\frac{1}{2}\right) - \sin \theta \left(\frac{\sqrt{3}}{2}\right) \]

Adding the three cosine terms:

\[ \cos \theta + \cos(\theta - 120^\circ) + \cos(\theta + 120^\circ) \]

\[ = \cos \theta - \frac{1}{2}\cos \theta + \frac{\sqrt{3}}{2}\sin \theta - \frac{1}{2}\cos \theta - \frac{\sqrt{3}}{2}\sin \theta \]

\[ = \cos \theta - \cos \theta \]

\[ = 0 \]

Thus, we have proved both identities.

\( \blacksquare \)

V. For any angles \( A \), \( B \), and \( C \), the following trigonometric identities hold:

The sine of the sum \( A + B + C \) can be expressed as:

\[ \sin (A + B + C) = \cos A \cos B \cos C (\tan A + \tan B + \tan C - \tan A \tan B \tan C) \]
The cosine of the sum \( A + B + C \) can be expressed as:

\[ \cos (A + B + C) = \cos A \cos B \cos C (1 - \tan B \tan C - \tan C \tan A - \tan A \tan B) \]

Proof:

For identity 1:

Using the sum of angles formula for sine, we have:

\[ \sin(A + B + C) = \sin(A + B) \cos C + \cos(A + B) \sin C \]

Applying the sum formula for \( \sin(A + B) \) and \( \cos(A + B) \), we get:

\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]

Substitute these into the original expression:

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

Grouping terms and factoring out \( \cos A \cos B \cos C \):

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

Recognizing that \( \frac{\sin \theta}{\cos \theta} = \tan \theta \), we substitute:

\[ = \cos A \cos B \cos C (\tan A + \tan B + \tan C - \tan A \tan B \tan C) \]

For identity 2:

Starting with the cosine of a sum formula:

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

Using the earlier mentioned formulas for \( \sin(A + B) \) and \( \cos(A + B) \), we substitute:

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

Grouping terms and factoring out \( \cos A \cos B \cos C \):

\[ = \cos A \cos B \cos C (1 - \frac{\sin B}{\cos B} \frac{\sin C}{\cos C} - \frac{\sin C}{\cos C} \frac{\sin A}{\cos A} - \frac{\sin A}{\cos A} \frac{\sin B}{\cos B}) \]

Again, recognizing the tangent function, we substitute:

\[ = \cos A \cos B \cos C (1 - \tan B \tan C - \tan C \tan A - \tan A \tan B) \]

Thus, we have proved both identities.

\( \blacksquare \)

VI. For any angles \( A \), \( B \), and \( C \), the following trigonometric identities hold:

The tangent of the sum \( A + B + C \) can be expressed as:

\[ \tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan B \tan C - \tan C \tan A - \tan A \tan B} \]
The cotangent of the sum \( A + B + C \) can be similarly expressed as:

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

Proof:

For identity 1:

Using the previously derived expressions for \( \sin(A + B + C) \) and \( \cos(A + B + C) \), and recalling that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we have:

\[ \tan(A + B + C) = \frac{\sin(A + B + C)}{\cos(A + B + C)} \]

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

Simplifying by dividing out the common factor of \( \cos A \cos B \cos C \) from numerator and denominator, we obtain:

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

For identity 2:

Starting with the cotangent identity \( \cot \theta = \frac{1}{\tan \theta} \), we express \( \cot(A + B + C) \) in terms of tangents:

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

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

Now, replacing \( \tan A \), \( \tan B \), and \( \tan C \) with \( \frac{1}{\cot A} \), \( \frac{1}{\cot B} \), and \( \frac{1}{\cot C} \) respectively, and multiplying numerator and denominator by \( \cot A \cot B \cot C \), we get:

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

Simplifying this expression, we obtain:

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

Thus, we have proved both identities.

\( \blacksquare \)

VII. For a set of \( n \) angles \( A_1, A_2, ..., A_n \), the sine and cosine of the sum of these angles can be expressed in terms of the sums of products of sines and cosines of these angles in the following way:

Let \( S_r \) represent the sum of products of the sines of \( r \) angles and the cosines of the remaining \( n - r \) angles from \( A_1, A_2, ..., A_n \), with every possible combination of \( r \) angles being considered.

Then, the sine of the sum of the angles is given by:

\[ \sin(A_1 + A_2 + ... + A_n) = S_1 - S_3 + S_5 - ... \]

and the cosine of the sum of the angles is given by:

\[ \cos(A_1 + A_2 + ... + A_n) = S_0 - S_2 + S_4 - ... \]

Where:

\[ S_0 = \cos A_1 \cos A_2 ... \cos A_n \]

\[ S_1 = \sin A_1 \cos A_2 ... \cos A_n + \cos A_1 \sin A_2 \cos A_3 ... \cos A_n + ... \]

The proof of this theorem can be constructed using mathematical induction, a powerful proof technique that will be discussed in a subsequent chapter.

Example

Let us apply it for sine and cosine of sum of four angles:

Using the theorem mentioned above, we can express \( \sin(A + B + C + D) \) and \( \cos(A + B + C + D) \) in terms of the sums of products of sines and cosines of the angles \( A, B, C, \) and \( D \).

For \( \sin(A + B + C + D) \), we will have the following terms:

\( S_1 \) will be the sum of the products of the sine of one angle and the cosines of the remaining three angles.
\( S_3 \) will be the sum of the products of the sines of three angles and the cosine of the remaining one angle.

Therefore, \( \sin(A + B + C + D) \) can be expressed as:

\[ \sin(A + B + C + D) = S_1 - S_3 \]

where

\[ S_1 = \sin A \cos B \cos C \cos D + \cos A \sin B \cos C \cos D + \cos A \cos B \sin C \cos D + \cos A \cos B \cos C \sin D \]

and

\[ S_3 = \sin A \sin B \sin C \cos D + \sin A \sin B \cos C \sin D + \sin A \cos B \sin C \sin D + \cos A \sin B \sin C \sin D \]

For \( \cos(A + B + C + D) \), we will have the following terms:

\( S_0 \) will be the product of the cosines of all four angles.
\( S_2 \) will be the sum of the products of the cosines of two angles and the sines of the remaining two angles.
\( S_4 \) will be the sum of the products of the sines of all four angles.

Therefore, \( \cos(A + B + C + D) \) can be expressed as:

\[ \cos(A + B + C + D) = S_0 - S_2 + S_4 \]

where

\[ S_0 = \cos A \cos B \cos C \cos D \]

\[ S_2 = \sin A \sin B \cos C \cos D + \sin A \cos B \sin C \cos D + \sin A \cos B \cos C \sin D + \cos A \sin B \sin C \cos D + \cos A \sin B \cos C \sin D + \cos A \cos B \sin C \sin D \]

\[ S_4 = \sin A \sin B \sin C \sin D \]

So the full expression for \( \cos(A + B + C + D) \) using the sums \( S_0 \), \( S_2 \), and \( S_4 \) is:

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

VIII. The tangent of the sum of \( n \) angles, \( A_1, A_2, ..., A_n \), can be expressed in terms of the tangents of those individual angles.

Let \( t_r \) denote the sum of the products of the tangents of \( r \) angles taken \( r \) at a time from \( A_1, A_2, ..., A_n \).

Then, the tangent of the sum of the \( n \) angles is given by the formula:

\[ \tan(A_1 + A_2 + ... + A_n) = \frac{t_1 - t_3 + t_5 - ...}{1 - t_2 + t_4 - ...} \]

Proof:

Starting from the known identities for the sine and cosine of the sum of \( n \) angles, we have:

\[ \sin(A_1 + A_2 + ... + A_n) = \cos A_1 \cos A_2 ... \cos A_n (t_1 - t_3 + t_5 - ...) \]

\[ \cos(A_1 + A_2 + ... + A_n) = \cos A_1 \cos A_2 ... \cos A_n (1 - t_2 + t_4 - ...) \]

Dividing the sine of the sum by the cosine of the sum, we obtain the tangent of the sum:

\[ \tan(A_1 + A_2 + ... + A_n) = \frac{\sin(A_1 + A_2 + ... + A_n)}{\cos(A_1 + A_2 + ... + A_n)} \]

Substituting the identities, we get:

\[ \tan(A_1 + A_2 + ... + A_n) = \frac{\cos A_1 \cos A_2 ... \cos A_n (t_1 - t_3 + t_5 - ...)}{\cos A_1 \cos A_2 ... \cos A_n (1 - t_2 + t_4 - ...)} \]

Since the products of cosines \( \cos A_1 \cos A_2 ... \cos A_n \) appear in both the numerator and the denominator, they cancel each other out, yielding:

\[ \tan(A_1 + A_2 + ... + A_n) = \frac{t_1 - t_3 + t_5 - ...}{1 - t_2 + t_4 - ...} \]

This completes the proof that the tangent of the sum of \( n \) angles can be expressed in terms of the sums \( t_1, t_2, t_3, ... \), which are the sums of products of tangents of those angles taken \( r \) at a time.

Q.E.D. (Quod Erat Demonstrandum)