Calculating Trigonometric Expression Value

The problem asks us to find the value of a specific trigonometric expression given the angles \(a = 45^\circ\) and \(b = 15^\circ\). The expression is:

$$ {\cos (a - b ) - \cos (a + b)} \over {\cos(a - b) + \cos (a + b)} $$

Substituting Angle Values

First, substitute the given values of \(a\) and \(b\) into the expression:

• \(a - b = 45^\circ - 15^\circ = 30^\circ\)
• \(a + b = 45^\circ + 15^\circ = 60^\circ\)

The expression becomes:

$$ {\cos (30^\circ) - \cos (60^\circ)} \over {\cos(30^\circ) + \cos (60^\circ)} $$

Evaluating Cosine Values

We need the standard values for \(\cos(30^\circ)\) and \(\cos(60^\circ)\). These are:

• \(\cos(30^\circ) = {\sqrt{3} \over 2}\)
• \(\cos(60^\circ) = {1 \over 2}\)

Substituting and Simplifying the Expression

Now, substitute these values into the expression:

$$ { {\sqrt{3} \over 2} - {1 \over 2} } \over { {\sqrt{3} \over 2} + {1 \over 2} } $$

Combine the terms in the numerator and the denominator:

$$ { {\sqrt{3} - 1} \over 2 } \over { {\sqrt{3} + 1} \over 2 } $$

To divide the fractions, multiply the numerator by the reciprocal of the denominator:

$$ { {\sqrt{3} - 1} \over 2 } \times { 2 \over {\sqrt{3} + 1} } $$

The '2' in the numerator and denominator cancel out:

$$ { {\sqrt{3} - 1} \over {\sqrt{3} + 1} } $$

Rationalizing the Denominator

To simplify further, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is \({\sqrt{3} - 1}\):

$$ { ({\sqrt{3} - 1}) \times ({\sqrt{3} - 1}) } \over { ({\sqrt{3} + 1}) \times ({\sqrt{3} - 1}) } $$

In the numerator, we use the formula \((x - y)^2 = x^2 - 2xy + y^2\). In the denominator, we use the formula \((x + y)(x - y) = x^2 - y^2\):

$$ { (\sqrt{3})^2 - 2(\sqrt{3})(1) + (1)^2 } \over { (\sqrt{3})^2 - (1)^2 } $$

$$ { 3 - 2\sqrt{3} + 1 } \over { 3 - 1 } $$

$$ { 4 - 2\sqrt{3} } \over { 2 } $$

Factor out 2 from the numerator:

$$ { 2 (2 - \sqrt{3}) } \over { 2 } $$

Cancel out the 2:

$$ 2 - \sqrt{3} $$

Alternative Method using Trigonometric Identities

The expression can be simplified using sum-to-product formulas or by recognizing a specific form related to the tangent function. Recall the formulas:

• \(\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\)
• \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\)

Let \(A = a-b\) and \(B = a+b\). Then \(\frac{A+B}{2} = \frac{(a-b) + (a+b)}{2} = \frac{2a}{2} = a\) and \(\frac{A-B}{2} = \frac{(a-b) - (a+b)}{2} = \frac{-2b}{2} = -b\).

So the numerator is \(\cos(a-b) - \cos(a+b) = -2 \sin(a) \sin(-b) = -2 \sin(a) (-\sin b) = 2 \sin a \sin b\).

The denominator is \(\cos(a-b) + \cos(a+b) = 2 \cos(a) \cos(-b) = 2 \cos a \cos b\) (since \(\cos(-x) = \cos x\)).

The expression becomes:

$$ { 2 \sin a \sin b } \over { 2 \cos a \cos b } = { \sin a \sin b } \over { \cos a \cos b } = \left({\sin a \over \cos a}\right) \left({\sin b \over \cos b}\right) = \tan a \tan b $$

Now substitute \(a = 45^\circ\) and \(b = 15^\circ\):

$$ \tan(45^\circ) \tan(15^\circ) $$

We know \(\tan(45^\circ) = 1\).

To find \(\tan(15^\circ)\), we use the tangent subtraction formula \(\tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\). Let \(x = 45^\circ\) and \(y = 30^\circ\):

$$ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = { \tan(45^\circ) - \tan(30^\circ) } \over { 1 + \tan(45^\circ) \tan(30^\circ) } $$

Substitute \(\tan(45^\circ) = 1\) and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):

$$ { 1 - {1 \over \sqrt{3}} } \over { 1 + 1 \times {1 \over \sqrt{3}} } = { { {\sqrt{3} - 1} \over \sqrt{3} } } \over { { {\sqrt{3} + 1} \over \sqrt{3} } } = { {\sqrt{3} - 1} \over \sqrt{3} } \times { \sqrt{3} \over {\sqrt{3} + 1} } = { {\sqrt{3} - 1} \over {\sqrt{3} + 1} } $$

Rationalizing this is the same as done in the first method, which gives \(2 - \sqrt{3}\).

So, the value of the expression is \(\tan(45^\circ) \tan(15^\circ) = 1 \times (2 - \sqrt{3}) = 2 - \sqrt{3}\).

Final Result

Both methods yield the same result. The value of the expression is \(2 - \sqrt{3}\).

Revision Table: Key Trigonometric Concepts

Additional Information on Trigonometry Calculations

This problem demonstrates how knowledge of basic trigonometric values and identities is crucial for simplifying expressions and solving problems. The identity \(\frac{\cos A - \cos B}{\cos A + \cos B} = \tan\left(\frac{A+B}{2}\right) \tan\left(\frac{B-A}{2}\right)\) or equivalently \(\tan\left(\frac{A+B}{2}\right) (-\tan\left(\frac{A-B}{2}\right))\) derived from the sum-to-product formulas, or the specific case \(\frac{\cos(x-y) - \cos(x+y)}{\cos(x-y) + \cos(x+y)} = \tan x \tan y\) used here, are very useful shortcuts. Knowing the values of trigonometric functions for common angles like 0°, 30°, 45°, 60°, and 90° is fundamental. For other angles like 15° or 75°, these values can often be derived using sum or difference formulas involving the common angles.