Problem 45 - Calculus Single Variable

Question

Use the identity \(\pi / 3+\pi / 4=7 \pi / 12\) to calculate the six trigonometric functions at \(7 \pi / 12\).

Step-by-Step Solution

Answer

At \(7\pi/12\): \(\cos = \frac{\sqrt{2} - \sqrt{6}}{4}\), \(\sin = \frac{\sqrt{6} + \sqrt{2}}{4}\), \(\tan = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{2} - \sqrt{6}}\).

Step 1: Understand the Identity

The identity given is \( \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12} \). This means that the angle \( \frac{7\pi}{12} \) can be expressed as the sum of two angles: \( \frac{\pi}{3} \) and \( \frac{\pi}{4} \). This allows us to use angle-sum identities to find the trigonometric functions at \( \frac{7\pi}{12} \).

Step 2: Calculate Cosine Using Angle-Sum Identity

The cosine of a sum is given by the identity:\[\cos(a + b) = \cos a \cos b - \sin a \sin b\]Substitute \( a = \frac{\pi}{3} \) and \( b = \frac{\pi}{4} \):\[\cos\left(\frac{7\pi}{12}\right) = \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right)\]\[\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\]\[\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\]Plug these into the identity:\[\cos\left(\frac{7\pi}{12}\right) = \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\]\[= \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}\]

Step 3: Calculate Sine Using Angle-Sum Identity

The sine of a sum is given by the identity:\[\sin(a + b) = \sin a \cos b + \cos a \sin b\]Substitute \( a = \frac{\pi}{3} \) and \( b = \frac{\pi}{4} \):\[\sin\left(\frac{7\pi}{12}\right) = \sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right)\]Using the known values:\[\sin\left(\frac{7\pi}{12}\right) = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\]\[= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\]

Step 4: Calculate Tangent Using Sine and Cosine

The tangent of an angle is the ratio of its sine to its cosine:\[\tan(a) = \frac{\sin(a)}{\cos(a)}\]For \( \frac{7\pi}{12} \):\[\tan\left(\frac{7\pi}{12}\right) = \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{2} - \sqrt{6}}{4}}\]Simplify:\[= \frac{\sqrt{6} + \sqrt{2}}{\sqrt{2} - \sqrt{6}}\]

Step 5: Find Reciprocal Functions

The reciprocal functions are derived from sine, cosine, and tangent:- Cosecant \( \csc(a) = \frac{1}{\sin(a)} \)- Secant \( \sec(a) = \frac{1}{\cos(a)} \)- Cotangent \( \cot(a) = \frac{1}{\tan(a)} \)For \( \frac{7\pi}{12} \):\[\csc\left(\frac{7\pi}{12}\right) = \frac{4}{\sqrt{6} + \sqrt{2}}, \quad \sec\left(\frac{7\pi}{12}\right) = \frac{4}{\sqrt{2} - \sqrt{6}}\]\[\cot\left(\frac{7\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{\sqrt{6} + \sqrt{2}}\]

Key Concepts

Angle-Sum Identities

Trigonometric functions are often calculated using angle-sum identities. This technique comes into play when dealing with angles expressed as the sum or difference of two others. In particular, the angle-sum identities for sine and cosine are:

• For sine: \( \sin(a + b) = \sin a \cos b + \cos a \sin b \)
• For cosine: \( \cos(a + b) = \cos a \cos b - \sin a \sin b \)

These identities are invaluable for finding the trigonometric functions of angles not located on the standard unit circle positions. For instance, if you have \( \frac{7\pi}{12} \), it can be broken down into \( \frac{\pi}{3} + \frac{\pi}{4} \). This dismantling makes it possible to determine sine and cosine values for \( \frac{7\pi}{12} \), based on already known values for \( \frac{\pi}{3} \) and \( \frac{\pi}{4} \). By utilizing these identities, you leverage the known angles to calculate new angle results easily. With practice, deciphering these types of identities becomes a quicker and more intuitive process.

Cosine

Cosine is a fundamental trigonometric function that relates the angle to the x-coordinate of a point on the unit circle. When calculating \( \cos(\frac{7\pi}{12}) \), instead of directly finding this value, we use the angle-sum identity. For the angle \( \frac{7\pi}{12} \), use:

• \( \cos(\frac{\pi}{3}) = \frac{1}{2} \)
• \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)

Apply the formula:\[\cos(\frac{7\pi}{12}) = \cos(\frac{\pi}{3})\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{3})\sin(\frac{\pi}{4})\]Substituting the known values results in: \( \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \). This expression simplifies to provide the precise value of \( \cos(\frac{7\pi}{12}) \). Understanding how to break down these expressions helps in memorizing and applying trigonometric values for angles that aren’t immediately familiar.

Sine

The sine function, similar to cosine, is one of the primary trigonometric functions. It connects an angle to the y-coordinate of a point on the unit circle. Knowing how to manipulate angle-sum identities allows for sine values of specific angles to be effortlessly computed. In the context of \( \frac{7\pi}{12} \), you utilize:

• \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \)
• \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)

This enables you to calculate:\[\sin(\frac{7\pi}{12}) = \sin(\frac{\pi}{3})\cos(\frac{\pi}{4}) + \cos(\frac{\pi}{3})\sin(\frac{\pi}{4})\]Substituting these into the equation gives: \( \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \). This outcome gives the sine value for \( \frac{7\pi}{12} \). Focusing on these processes helps solidify understanding of how sine behaves in different segments of the unit circle.

Tangent

Tangent, another essential trigonometric function, is the ratio of sine to cosine. Often, evaluating tangent directly is less feasible compared to breaking down this ratio using known sine and cosine values. For \( \tan(\frac{7\pi}{12}) \), leverage the previously calculated sine and cosine:

• \( \sin(\frac{7\pi}{12}) = \frac{\sqrt{6} + \sqrt{2}}{4} \)
• \( \cos(\frac{7\pi}{12}) = \frac{\sqrt{2} - \sqrt{6}}{4} \)

Use the formula:\[\tan(\frac{7\pi}{12}) = \frac{\sin(\frac{7\pi}{12})}{\cos(\frac{7\pi}{12})}\]This simplifies to \( \frac{\sqrt{6} + \sqrt{2}}{\sqrt{2} - \sqrt{6}} \), which, though complicated, highlights the nature of tangent derived from angle sums. Appropriately handling these calculations ensures an accurate grasp of tangent's relationship to sine and cosine.

Reciprocal Trigonometric Functions

The reciprocal trigonometric functions—which include cosecant, secant, and cotangent—are inverses of sine, cosine, and tangent, respectively. They offer alternatives for solving various trigonometric problems, especially when dealing with specific values or angles. Given:

• \( \csc(a) = \frac{1}{\sin(a)} \)
• \( \sec(a) = \frac{1}{\cos(a)} \)
• \( \cot(a) = \frac{1}{\tan(a)} \)

For the angle \( \frac{7\pi}{12} \), apply these reciprocations:

• \( \csc(\frac{7\pi}{12}) = \frac{4}{\sqrt{6} + \sqrt{2}} \)
• \( \sec(\frac{7\pi}{12}) = \frac{4}{\sqrt{2} - \sqrt{6}} \)
• \( \cot(\frac{7\pi}{12}) = \frac{\sqrt{2} - \sqrt{6}}{\sqrt{6} + \sqrt{2}} \)

Understanding how these reciprocals interact with angle-sum identities further assists solving trigonometric equations efficiently. Mastering these relationships creates additional pathways to tackle more elaborate trigonometric scenarios.