Problem 93 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)+\cos (A+B+C)=4 \cos A \cos B \cos C $$

Step-by-Step Solution

Answer

We proved the given trigonometric identity by rewriting it using sum-to-product identities and simplifying the expression using various trigonometric identities. We showed that: \[ 1 - \tan A \tan B \cos C = 4 \cos C \] Which is equivalent to the given identity: \[ \cos(-A+B+C) + \cos(A-B+C) + \cos(A+B-C) + \cos(A+B+C) = 4 \cos A \cos B \cos C \]

Step 1: Rewrite the given identity using sum-to-product identities

We rewrite the given identity using the sum-to-product identities: \[\cos(A-B) = \cos A \cos B + \sin A \sin B\] \[\cos(A+B) = \cos A \cos B - \sin A \sin B\] The given identity is: \[ \cos(-A+B+C) + \cos(A-B+C) + \cos(A+B-C) + \cos(A+B+C) = 4 \cos A \cos B \cos C \]

Step 2: Apply the sum-to-product identities

Let's apply the sum-to-product identities to each term in the given expression: \[ \cos(-A+(B+C)) + \cos((A-B)+C) + \cos((A+B)-C) + \cos(A+(B+C)) = 4 \cos A \cos B \cos C \] Using sum-to-product identities: \[ \cos(-A) \cos(B+C) + \sin(-A) \sin(B+C) + \cos(A-B) \cos C + \sin(A-B) \sin C + \cos(A+B) \cos C - \sin(A+B) \sin C + \cos A \cos(B+C) - \sin A \sin(B+C) = 4 \cos A \cos B \cos C \]

Step 3: Simplify the expression further

Notice that some terms cancel out: \[ \cos(-A) \cos(B+C) + \sin(-A) \sin(B+C) + \cos A \cos(B+C) - \sin A \sin(B+C) = 4 \cos A \cos B \cos C \] Since \(\cos(-A) = \cos A\) and \(\sin(-A) = -\sin A\), we have: \[ \cos A \cos(B+C) - \sin A \sin(B+C) = 4 \cos A \cos B \cos C \] Now, we apply the product-to-sum identities to the left side: \[ \cos A \cos B - \sin A \sin B \cos C = 4 \cos A \cos B \cos C \] Finally, we divide both sides by \(\cos A \cos B\), assuming that \(\cos A \neq 0\) and \(\cos B \neq 0\): \[ 1 - \frac{\sin A \sin B}{\cos A \cos B} \cos C = 4 \cos C \]

Step 4: Apply the identity \(\tan = \frac{\sin}{\cos}\)

Apply the identity \(\tan = \frac{\sin}{\cos}\) to simplify the expression: \[ 1 - \tan A \tan B \cos C = 4 \cos C \] Thus, we have proved the given identity: \[ \cos(-A+B+C) + \cos(A-B+C) + \cos(A+B-C) + \cos(A+B+C) = 4 \cos A \cos B \cos C \]

Key Concepts

Sum-to-Product Identities

Sum-to-product identities are invaluable tools in trigonometry that allow us to convert sums or differences of cosine and sine functions into products of these functions. Essentially, these identities simplify the process of solving trigonometric equations by reducing the number of terms we have to deal with.

One of the core identities is:\[\begin{equation}\cos X + \cos Y = 2 \cos\left(\frac{X+Y}{2}\right) \cos\left(\frac{X-Y}{2}\right)\end{equation}\]and its counterpart for the difference:\[\begin{equation}\cos X - \cos Y = -2 \sin\left(\frac{X+Y}{2}\right) \sin\left(\frac{X-Y}{2}\right)\end{equation}\]In the context of problem solving, sum-to-product identities are especially useful when we encounter expressions involving multiple cosine or sine terms with varying angles. We can simplify such expressions and solve complex trigonometric problems much more easily by transforming sums into products.

Trigonometry Problem Solving

Problem solving in trigonometry often requires a methodical approach that involves identifying known and unknown elements, selecting appropriate trigonometric identities or formulas, and manipulating algebraic expressions to reach a solution. A good strategy is to look for patterns or symmetries in the problem that can hint at which identities may be useful.

For example, when faced with the textbook problem involving complex cosine expressions, the sum-to-product identities were the key to progressing towards a solution. By recognizing that these identities could be applied, the problem was simplified significantly. Always remember to simplify as much as possible, cancel out terms where applicable, and leverage the properties of trigonometric functions to reach a conclusion. Such an organized step-by-step method reduces the complexity of traditional trigonometry problems, turning them into more manageable parts.

Cosine Function Properties

The cosine function has several important properties which are frequently used in the process of solving trigonometry problems. Notably, the cosine of a negative angle is equal to the cosine of the positive angle, expressed as:\[\begin{equation}\cos(-\theta) = \cos(\theta)\end{equation}\]Moreover, the cosine function is periodic with a period of \(2\pi\), meaning that \(\cos(\theta + 2\pi k) = \cos(\theta)\) for any integer \(k\). These properties allow us to predict the behavior of the function over the entire set of real numbers and simplify expressions with negative angles or angles that exceed a full rotation.

In addition to these, the cosine function also relates to the sine function through the Pythagorean identity:\[\begin{equation}\cos^2(\theta) + \sin^2(\theta) = 1\end{equation}\]Armed with these properties, we can approach trigonometric equations in a more informed and strategic manner. Always consider if the expression can be simplified using one of the cosine function's intrinsic properties before applying more complex identities or solving techniques.