Problem 29 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\cos (x-y)=-1\), then prove that \(\cos x+\cos y=0\) and \(\sin x+\sin y=0\)

Step-by-Step Solution

Answer

\(\cos x + \cos y = 0\) and \(\sin x + \sin y = 0\).

Step 1: Invoke the Cosine of Difference Formula

Use the trigonometric identity for the cosine of the difference of two angles. The identity is \(\cos(a - b) = \cos a \cos b + \sin a \sin b\). Here, \(a = x\) and \(b = y\), so it becomes \(\cos (x - y) = \cos x \cos y + \sin x \sin y\).

Step 2: Substitute given

Now we substitute the given \(\cos (x - y) = -1\). By substituting this in the equation obtained from step 1, we get: \(-1 = \cos x \cos y + \sin x \sin y\).

Step 3: Derive \(\cos x + \cos y\)

By rearranging the above equation, we obtain \(\cos x \cos y = -1 - \sin x \sin y\). Since trigonometric identities says that \(\cos^2 A + \sin^2 A = 1\), it implies that \(\cos y = -\cos x\). Hence the sum of \(\cos x + \cos y = 0\).

Step 4: Derive \(\sin x + \sin y\)

From the identity \(\cos^2 A + \sin^2 A = 1\), substituting \(\cos y = -\cos x\), we get \(\sin y = \sin x\). Hence the sum of \(\sin x + \sin y = 0\).

Key Concepts

Cosine of Difference Formula

The Cosine of Difference Formula is a fundamental trigonometric identity used to calculate the cosine of the difference between two angles. It states that for any angles \(a\) and \(b\), the identity is:

• \(\cos(a - b) = \cos a \cos b + \sin a \sin b\)

This formula is particularly useful when solving trigonometric equations where differences of angles appear. Applying this formula helps express the cosine of two subtracted angles in terms of the sines and cosines of the angles themselves. When you replace \(a\) and \(b\) with specific values, as seen in the exercise where \(x\) and \(y\) represent angles, it transforms into:

• \(\cos(x-y) = \cos x \cos y + \sin x \sin y\)

Substituting \(\cos(x-y) = -1\) allows you to explore further relationships between the angles \(x\) and \(y\), such as proving conditions on the sum of their sine and cosine components.

Trigonometric Proof

Trigonometric proofs involve demonstrating the truth of a statement or equation using basic trigonometric identities and algebra. In this exercise, you need to prove the conditions \(\cos x + \cos y = 0\) and \(\sin x + \sin y = 0\) given that \(\cos(x-y) = -1\). Starting with the identity \(\cos(x - y) = \cos x \cos y + \sin x \sin y\) from the Cosine of Difference Formula, substitute \(-1\) for \(\cos(x-y)\):

• \(-1 = \cos x \cos y + \sin x \sin y\)

Rearrange to show relations between the cosine and sine components:

• \(\cos x \cos y = -1 - \sin x \sin y\)

Using the Pythagorean identity \(\cos^2 A + \sin^2 A = 1\), solve for \(\cos y\) and \(\sin y\) in terms of \(\cos x\) and \(\sin x\), resulting in the conclusions that the sums of the cosines and sines are zero.

Sine and Cosine Relationship

The relationship between sine and cosine is closely associated with the fundamental Pythagorean identity:

• \(\cos^2 A + \sin^2 A = 1\)

This identity provides a powerful connection between the sine and cosine of any angle. In the context of this problem, knowing that \(\cos(x)\) and \(\sin(x)\) are connected through such identities helps derive other relationships.After deducing from the formula \(\cos(x-y) = -1\), you find \(\cos y = -\cos x\) and \(\sin y = \sin x\). These relationships reveal:

• The sum \(\cos x + \cos y = 0\), since \(\cos y = -\cos x\).
• The sum \(\sin x + \sin y = 0\), as \(\sin y = \sin x\) which implies \(\sin y = -\sin x\) due to the identity \(\sin^2 y + \cos^2 y = 1\).

Using these relationships helps solidify the understanding of how sine and cosine work together to satisfy trigonometric identities and derive new proofs.