Solving Trigonometry: Finding Cosine of Angle Difference

The problem asks us to find the value of \(\cos (\alpha - \beta)\) given the values of \(\cos (\theta - \alpha)\) and \(\cos (\theta - \beta)\). This requires using fundamental trigonometric identities, specifically the angle difference formula for cosine.

We are given:

\(\cos (\theta - \alpha) = a\)

\(\cos (\theta - \beta) = b\)

We need to find \(\cos (\alpha - \beta)\).

Let's establish a connection between the given angles and the angle whose cosine we need to find. Notice that the difference \((\alpha - \beta)\) can be related to the given angles \((\theta - \alpha)\) and \((\theta - \beta)\).

Consider the difference between the two given angles:

$$(\theta - \beta) - (\theta - \alpha) = \theta - \beta - \theta + \alpha = \alpha - \beta$$

So, we can write \(\cos (\alpha - \beta)\) as \(\cos ((\theta - \beta) - (\theta - \alpha))\).

Let \(A = \theta - \alpha\) and \(B = \theta - \beta\). Then we are given \(\cos A = a\) and \(\cos B = b\), and we need to find \(\cos(B - A)\).

We use the cosine angle difference formula, which states that \(\cos(X - Y) = \cos X \cos Y + \sin X \sin Y\).

Applying this formula to \(\cos(B - A)\): $$\cos(B - A) = \cos B \cos A + \sin B \sin A$$

We know \(\cos A = a\) and \(\cos B = b\). Now we need to find \(\sin A\) and \(\sin B\) in terms of \(a\) and \(b\).

Using the fundamental trigonometric identity \(\sin^2 x + \cos^2 x = 1\), we can express \(\sin x\) as \(\sin x = \pm \sqrt{1 - \cos^2 x}\).

So, for angle \(A\):

$$\sin A = \sin(\theta - \alpha) = \pm \sqrt{1 - \cos^2 (\theta - \alpha)} = \pm \sqrt{1 - a^2}$$

And for angle \(B\):

$$\sin B = \sin(\theta - \beta) = \pm \sqrt{1 - \cos^2 (\theta - \beta)} = \pm \sqrt{1 - b^2}$$

Now substitute these values back into the formula for \(\cos(B - A)\): $$\cos(\alpha - \beta) = \cos(B - A) = (b)(a) + (\pm \sqrt{1 - b^2})(\pm \sqrt{1 - a^2})$$ $$\cos(\alpha - \beta) = ab + (\pm \sqrt{1 - a^2})(\pm \sqrt{1 - b^2})$$

The product of the two \(\pm\) terms can be either positive or negative. However, looking at the options provided in the question, the term involving the square roots is \(\sqrt{1 - a^2}\sqrt{1 - b^2}\) with a positive sign. This indicates that we should consider the case where \(\sin A\) and \(\sin B\) have the same sign (either both positive or both negative), resulting in a positive product \(\sin A \sin B = \sqrt{1 - a^2}\sqrt{1 - b^2}\).

Thus, based on the structure of the options, the expression for \(\cos(\alpha - \beta)\) is:

$$\cos(\alpha - \beta) = ab + \sqrt{1 - a^2}\sqrt{1 - b^2}$$

Let's check the options:

\(\(ab + \sqrt {1 - {a^2}} \;\sqrt {1 - b2}\)\) - Matches our derived expression. Note the slight typo in the original option text \(b2\) instead of \(b^2\). We assume it means \(b^2\).

\(\(ab = \sqrt {1 - {a^2}} \;\sqrt {1 - {b^2}}\)\) - This is an equation, not an expression for \(\cos(\alpha - \beta)\). Also contains the equal sign typo.

\(\(a\sqrt {1 - {b^2}} - b\sqrt {1 - {a^2}}\)\) - Does not match.

\(\(a\sqrt {1 - {b^2}} + b\sqrt {1 - {a^2}}\)\) - Does not match.

The first option, interpreting \(b2\) as \(b^2\), matches our result.

The final answer is \(ab + \sqrt{1 - a^2}\sqrt{1 - b^2}\).

Revision Table: Key Trigonometry Concepts

Understanding the following trigonometric identities is crucial for solving problems like this:

Pythagorean Identity: \(\sin^2 x + \cos^2 x = 1\). This allows finding sine from cosine (or vice versa).

Cosine Difference Formula: \(\cos(X - Y) = \cos X \cos Y + \sin X \sin Y\). This is used to expand or simplify the cosine of the difference of two angles.

Angle Manipulation: Expressing the desired angle \((\alpha - \beta)\) in terms of the given angles \(((\theta - \beta) - (\theta - \alpha))\) is a key step.

Remember that when taking the square root of \(1 - \cos^2 x\), the result is \(\pm \sqrt{1 - \cos^2 x}\). The specific context or the options provided in a question often guide which sign combination is relevant.

Additional Information: Sine and Cosine Relationships

The relationship between sine and cosine values for any angle \(x\) is given by the Pythagorean identity. This identity arises directly from the unit circle definition of trigonometric functions, where \(\cos x\) is the x-coordinate and \(\sin x\) is the y-coordinate of a point on the unit circle corresponding to angle \(x\). The equation of the unit circle is \(x^2 + y^2 = 1\), which translates to \(\cos^2 x + \sin^2 x = 1\).

The angle addition and subtraction formulas for cosine and sine are derived using geometric proofs or Euler's formula. The cosine difference formula, \(\cos(X - Y) = \cos X \cos Y + \sin X \sin Y\), is a fundamental identity. From this, other identities like \(\cos(X + Y)\), \(\sin(X - Y)\), and \(\sin(X + Y)\) can be derived. For example, \(\cos(X + Y) = \cos(X - (-Y)) = \cos X \cos(-Y) + \sin X \sin(-Y)\). Since \(\cos(-Y) = \cos Y\) and \(\sin(-Y) = -\sin Y\), we get \(\cos(X + Y) = \cos X \cos Y - \sin X \sin Y\).

In this problem, the step involving \(\sin(\theta - \alpha) = \pm \sqrt{1 - a^2}\) and \(\sin(\theta - \beta) = \pm \sqrt{1 - b^2}\) introduces ambiguity. However, the formula for \(\cos(B - A)\) requires the product \(\sin B \sin A\). The product of two values each with a \(\pm\) sign results in a \(\pm\) product. But the structure of the options guides us to select the positive product \(\sqrt{1 - a^2}\sqrt{1 - b^2}\), which implies \(\sin(\theta - \alpha)\) and \(\sin(\theta - \beta)\) have the same sign.