Let sin x + sin y = cos x + cos y for all x, y ∈ ℝ. What is \(\rm tan \left(\frac{x}{2}+\frac{y}{2}\right)\) equal to?
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Solving Trigonometric Identities: Finding tan((x+y)/2)
The problem asks us to find the value of \(\tan \left(\frac{x}{2}+\frac{y}{2}\right)\) given the trigonometric equation \(\sin x + \sin y = \cos x + \cos y\), which holds for all real numbers x and y.
To solve this, we will use sum-to-product trigonometric identities to simplify the given equation.
Applying Sum-to-Product Formulas
The given equation is:
\[\sin x + \sin y = \cos x + \cos y\]We use the following identities:
- \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\)
- \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\)
Applying these identities to the equation, we get:
\[2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)\]Simplifying the Trigonometric Equation
Now, we simplify the equation obtained from applying the identities. Divide both sides by 2:
\[\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)\]Move all terms to one side:
\[\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = 0\]Factor out the common term \(\cos\left(\frac{x-y}{2}\right)\):
\[\cos\left(\frac{x-y}{2}\right) \left[ \sin\left(\frac{x+y}{2}\right) - \cos\left(\frac{x+y}{2}\right) \right] = 0\]This equation holds true if either factor is equal to zero:
- \(\cos\left(\frac{x-y}{2}\right) = 0\)
- \(\sin\left(\frac{x+y}{2}\right) - \cos\left(\frac{x+y}{2}\right) = 0\)
The problem states that the original equation \(\sin x + \sin y = \cos x + \cos y\) holds for all \(x, y \in \mathbb{R}\). If \(\cos\left(\frac{x-y}{2}\right) = 0\) were the only condition, it would not be true for all x and y (for example, when x=y, \(\cos(0)=1 \neq 0\)). Therefore, the second condition must be the one that ensures the equality holds for all x and y.
\[\sin\left(\frac{x+y}{2}\right) - \cos\left(\frac{x+y}{2}\right) = 0\] \[\sin\left(\frac{x+y}{2}\right) = \cos\left(\frac{x+y}{2}\right)\]Finding the Value of tan((x+y)/2)
We have \(\sin\left(\frac{x+y}{2}\right) = \cos\left(\frac{x+y}{2}\right)\). Assuming \(\cos\left(\frac{x+y}{2}\right) \neq 0\), we can divide both sides by \(\cos\left(\frac{x+y}{2}\right)\) to find the value of the tangent.
If \(\cos\left(\frac{x+y}{2}\right) = 0\), then from the equation \(\sin\left(\frac{x+y}{2}\right) = \cos\left(\frac{x+y}{2}\right)\), it would imply \(\sin\left(\frac{x+y}{2}\right) = 0\). However, the sine and cosine of the same angle cannot both be zero simultaneously, as \(\sin^2 \theta + \cos^2 \theta = 1\). Thus, \(\cos\left(\frac{x+y}{2}\right)\) cannot be zero under this condition, and the division is valid.
\[\frac{\sin\left(\frac{x+y}{2}\right)}{\cos\left(\frac{x+y}{2}\right)} = 1\]By the definition of tangent, \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Therefore:
\[\tan\left(\frac{x+y}{2}\right) = 1\]The value of \(\tan \left(\frac{x}{2}+\frac{y}{2}\right)\) is 1.
Revision Table: Key Concepts
| Concept | Description | Identity/Formula |
|---|---|---|
| Sum of Sines | Expresses the sum of two sines as a product. | \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\) |
| Sum of Cosines | Expresses the sum of two cosines as a product. | \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\) |
| Tangent Identity | Relates sine and cosine for a given angle. | \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) |
Additional Information: Trigonometric Identities
Trigonometric identities are equalities involving trigonometric functions that hold true for all values of the variables for which the functions are defined. They are fundamental tools for simplifying expressions, solving equations, and proving other relationships in trigonometry.
Sum-to-product and product-to-sum identities are particularly useful for transforming sums or differences of sines and cosines into products, and vice versa. This transformation often helps in solving equations or evaluating expressions that are difficult in their original form.
For instance, the identity \(\sin \theta = \cos \theta\) implies \(\tan \theta = 1\), which means \(\theta\) must be of the form \(\frac{\pi}{4} + n\pi\), where \(n\) is an integer. In our case, \(\frac{x+y}{2}\) must be of this form.
Understanding and recognizing when to apply these identities is crucial for success in trigonometry and related fields.
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