Problem 41 - Algebra and Trigonometry with Analytic Geometry

Question

Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$ \sin \frac{1}{2} u+\cos u=1 $$

Step-by-Step Solution

Answer

The solution is \( u = \pi \).

Step 1: Substitute the given functions

Let's begin by introducing a substitution. Let \( v = \frac{1}{2}u \), so \( u = 2v \). Then our equation becomes \( \sin v + \cos(2v) = 1 \).

Step 2: Use trigonometric identities

Use the double angle identity for cosine: \( \cos(2v) = 2\cos^2(v) - 1 \). Substitute this into the equation: \( \sin v + (2\cos^2(v) - 1) = 1 \).

Step 3: Simplify the equation

Rearrange and group like terms: \( \sin v + 2\cos^2(v) = 2 \). Recall that \( \sin^2(v) + \cos^2(v) = 1 \), thus \( \sin^2(v) = 1 - \cos^2(v) \).

Step 4: Substitute using Pythagorean identity

Substitute \( \sin(v) = \sqrt{1 - \cos^2(v)} \) into the equation: \( \sqrt{1 - \cos^2(v)} + 2\cos^2(v) = 2 \).

Step 5: Solve the trigonometric equation

Assume \( \cos(v) = x \). Substitute: \( \sqrt{1 - x^2} + 2x^2 = 2 \). Since this is difficult to solve, test simple values where sine and cosine achieve values that satisfy the equation.

Step 6: Assume practical trigonometric solutions

Try typical angle values such as \( v = 0 \), \( v = \frac{\pi}{2} \), where trigonometric functions have exact values. At \( v = \frac{\pi}{3} \), we get \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and equivalently for cosines.

Step 7: Check and express solution in terms of original variable

We found that when \( v = \frac{\pi}{2} \), \( u = 2 \times \frac{\pi}{2} = \pi \). Verify that this satisfies the original equation.

Step 8: Include other solutions in the interval

Consider trigonometric periodicity. \( \sin(v + 2\pi k) = \sin(v) \) and \( \cos(v + 2\pi k) = \cos(v) \). In the given interval, the only solution derived is \( u = \pi \).

Key Concepts

Trigonometric Identities

Trigonometric identities are tools that help us manipulate and solve trigonometric equations. These identities relate the trigonometric functions to one another, allowing for simplifications and transformations.
One primary form of identity is the Pythagorean identity which includes key relationships like

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

You can also have angle sum and difference identities and double angle identities like

• \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \)
• \( \tan(\theta + \phi) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta)\tan(\phi)} \)

These identities help simplify and solve equations by rewriting trigonometric expressions in a more convenient form. In the given exercise, trigonometric identities were crucial in transforming the equation \( \sin \left(\frac{1}{2} u\right) + \cos(u) = 1 \) into a solvable form.

Pythagorean Identity

The Pythagorean identity is one of the most fundamental identities in trigonometry and is derived from the Pythagorean theorem in geometry. This identity states that for any angle \( \theta \), the square of the sine of the angle plus the square of the cosine of the angle is always equal to one:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

This identity is very useful for transforming one trigonometric function into another and can help with the simplification process of equations. In the example exercise, this concept was used to help substitute \( \sin^2(v)\) with \(1 - \cos^2(v)\), thereby allowing a more straightforward simplification of the equation \( \sin(v) + 2\cos^2(v) = 2 \). Recognizing and applying the Pythagorean identity is a powerful method for solving trigonometric expressions by effectively reducing the equation to a simpler form.

Double Angle Identity

The double angle identity is incredibly useful when dealing with equations that involve angles that are double or half a given angle, such as in our exercise where \( u \) and \( \frac{1}{2} u \) are present. The double angle identity particularly studied here is for cosine:

• \( \cos(2\theta) = 2\cos^2(\theta) - 1 \)

This identity allows us to express the cosine of a double angle in terms of a single angle function. In the solved problem, the identity was applied to transform \( \cos(2v) \) into a form involving only squared cosine, \( 2\cos^2(v) - 1 \), which aids significantly in simplifying the trigonometric equation towards a solvable form. By using such identities, complex trigonometric expressions can become much more manageable and straightforward to solve.

Interval Solutions

Finding solutions within a specified interval is a common requirement in trigonometry, particularly in exams and quizzes. This requirement involves finding the specific values of the solution that lie within a given range, typically between 0 and \(2\pi\) for trigonometric equations.
The exercise placed this restriction by asking for solutions for \( u \) where \( 0 \leq u < 2\pi \). While solving, it's essential to remain aware of these boundaries and disregard any solutions that fall outside.Checking periodicity is also crucial due to the nature of trigonometric functions, which are cyclical. Once a solution is found, as was the case where \( u = \pi \), additional solutions might fit by understanding the periodic nature of sine and cosine functions (e.g., repeating solutions every \(2\pi\) radians). Utilizing intervals and periodicity principles ensures that only valid solutions are identified within the specified range.