Problem 51
Question
Solve the equation for \(x\) in terms of \(y\) if \(0 < x < \pi\) and \(0 < y < \pi\). $$\frac{\sin x}{3}=\frac{\sin y}{4}$$
Step-by-Step Solution
Verified Answer
\( x = \sin^{-1}\left(\frac{3}{4} \sin y\right) \) for \(0 < x < \pi\).
1Step 1: Understand the Equation
The given equation is \( \frac{\sin x}{3} = \frac{\sin y}{4} \). This is a proportion that relates \( \sin x \) to \( \sin y \) through the fractions \( \frac{1}{3} \) and \( \frac{1}{4} \).
2Step 2: Cross-Multiply to Eliminate Fractions
Multiply both sides of the equation by 12 (which is the least common multiple of 3 and 4) to clear the denominators:\[4 \sin x = 3 \sin y\]
3Step 3: Solve for \(\sin x\)
Isolate \( \sin x \) on one side by dividing both sides of the equation by 4:\[\sin x = \frac{3}{4} \sin y\]
4Step 4: Solve for \(x\) in terms of \(y\)
To find \( x \), take the inverse sine (arcsine) of both sides:\[x = \sin^{-1}\left(\frac{3}{4} \sin y\right)\]Since \( 0 < x < \pi \), this solution is valid as long as \( \frac{3}{4} \sin y \leq 1 \).
Key Concepts
Inverse Trigonometric FunctionsTrigonometric IdentitiesAngle Constraints
Inverse Trigonometric Functions
When solving trigonometric equations, we often need to employ inverse trigonometric functions like arcsine, arccosine, or arctangent. These functions help us find the angle whose trigonometric value we already know. In our original problem, we used arcsine, denoted as \( \sin^{-1} \), to solve for \( x \).
The equation \( x = \sin^{-1}\left(\frac{3}{4} \sin y\right) \) requires finding the angle \( x \) such that \( \sin(x) = \frac{3}{4} \sin(y) \). This process involves:
The equation \( x = \sin^{-1}\left(\frac{3}{4} \sin y\right) \) requires finding the angle \( x \) such that \( \sin(x) = \frac{3}{4} \sin(y) \). This process involves:
- Calculating the inverse sine of a value
- Considering only the principal value, which lies between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
Trigonometric Identities
Trigonometric identities are equations that are true for all values of the involved variables. They are crucial in simplifying trigonometric equations. In the given equation, we used the basic idea of proportions rather than a specific identity. However, identities can also offer shortcuts or alternative approaches to solving trig equations.
Some fundamental trigonometric identities you might encounter include:
Some fundamental trigonometric identities you might encounter include:
- Pythagorean identities like \(\sin^2\theta + \cos^2\theta = 1\)
- Angle sum and difference identities
- Double angle identities such as \(\sin(2\theta) = 2\sin\theta\cos\theta\)
Angle Constraints
In trigonometry problems, knowing the constraints on angles is essential. These constraints often determine the validity of solutions and keep them within expected bounds.
For the equation \(\frac{\sin x}{3} = \frac{\sin y}{4}\), the constraints given were \(0 < x < \pi\) and \(0 < y < \pi\). These specify that both \(x\) and \(y\) must be positive angles and not exceed \(\pi\) (or 180 degrees). This range is typical when dealing with sine functions because it covers all possible positive values of sine.
Therefore, when you solve \(x = \sin^{-1}\left(\frac{3}{4} \sin y\right)\), you need to ensure \( \frac{3}{4}\sin y \leq 1 \) to avoid ending up with an undefined or out-of-bound solution. These angle constraints, hence, help us filter out feasible solutions.
For the equation \(\frac{\sin x}{3} = \frac{\sin y}{4}\), the constraints given were \(0 < x < \pi\) and \(0 < y < \pi\). These specify that both \(x\) and \(y\) must be positive angles and not exceed \(\pi\) (or 180 degrees). This range is typical when dealing with sine functions because it covers all possible positive values of sine.
Therefore, when you solve \(x = \sin^{-1}\left(\frac{3}{4} \sin y\right)\), you need to ensure \( \frac{3}{4}\sin y \leq 1 \) to avoid ending up with an undefined or out-of-bound solution. These angle constraints, hence, help us filter out feasible solutions.
Other exercises in this chapter
Problem 50
If \(\alpha\) and \(\beta\) are complementary angles, show that $$ \sin ^{2} \alpha+\sin ^{2} \beta=1 $$
View solution Problem 50
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$2 \sin ^{2} u+\sin u-6=0$$
View solution Problem 51
Show that the equation is not an Identity. $$\cos t=\sqrt{1-\sin ^{2} t}$$
View solution Problem 51
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$1-\sin t=\sqrt{3} \cos t$$
View solution