Question

If \(5 \mathrm{x}+12 \mathrm{y}+13=0\) is transformed into \(x \cos \alpha+\mathrm{ysin} \alpha=\mathrm{p}\) then \(\alpha=? \alpha \in[-\pi, \pi]\) (a) \(\cos ^{-1}[-(5 / 13)]\) (b) \(\sin ^{-1}[-(12 / 13)]\) (c) \(\tan ^{-1}(12 / 5)-\pi\) (d) \(\tan ^{-1}(12 / 5)\)

Short answer

(a) \(\cos^{-1}\left[-\left(\frac{5}{13}\right)\right]\)

Step-by-step solution

Analyzing the given information

We have the linear equation, \(5x + 12y + 13 = 0\), and its transformed equation, \(x\cos\alpha + y\sin\alpha = p\). We need to find the relationship between the coefficients of the given equation and the transformed equation.

Make the given equation similar to the transformed one

Dividing the entire equation by \(\sqrt{(5^2 + 12^2)}\), or \(\sqrt{169}\), will give us the equation in a form resembling the transformed equation: \(\frac{5}{13}x + \frac{12}{13}y + \frac{13}{13} = 0\) Now it looks similar to \(x\cos\alpha + y\sin\alpha = p\).

Comparing coefficients

We can now see that there is a relationship between the coefficients of the two equations: \(\frac{5}{13} = \cos\alpha\) ⇒ \(\cos\alpha = \frac{5}{13}\) \(\frac{12}{13} = \sin\alpha\) ⇒ \(\sin\alpha = \frac{12}{13}\) The next step is to use these relationships to find \(\alpha\).

Finding α using inverse trigonometric functions

We can use the inverse trigonometric functions to find \(\alpha\). We have the following options: (a) \(\cos ^{-1}[-(5 / 13)]\) (b) \(\sin ^{-1}[-(12 / 13)]\) (c) \(\tan ^{-1}(12 / 5)-\pi\) (d) \(\tan ^{-1}(12 / 5)\) However, these do not directly match with what we have found: \(\cos\alpha = \frac{5}{13}\) and \(\sin\alpha = \frac{12}{13}\). We must first recognize that the coefficients of the original equation have opposite signs compared to the transformed equation. This means we're looking for answers that use the negative of these coefficients—options (a) and (b).

Finding the correct answer

We can now choose the correct answer from our choices (a) and (b). Since we've accounted for the negative sign, \(\cos^{-1}(5/13)\) and \(\sin^{-1}(12/13)\) should give the same value of \(\alpha\). Thus, the correct answer is: (a) \(\cos^{-1}\left[-\left(\frac{5}{13}\right)\right]\)

Key concepts

Linear Equations

Linear equations are fundamental in algebra and represent the relationship between two variables in the form of a straight line when plotted on a graph. They typically take the form of ax + by + c = 0, where a, b, and c are coefficients and constants, and x and y are variables.

Understanding how to manipulate and solve linear equations is crucial, as they appear frequently in various areas of mathematics and the sciences. The process often involves isolating one variable, simplifying algebraic expressions, and solving for the value of the variable. One way to transform a linear equation is by normalizing it, which can be done by dividing the entire equation by a common factor, making the comparison to another form clearer. This method is especially useful when comparing coefficients in different representations of a linear equation, such as the example in our exercise.

Trigonometric Coefficients

In trigonometry, the coefficients of trigonometric functions like sin and cos can be seen as instructions on how to scale or adjust the amplitude of these functions. Trigonometric coefficients within linear equations can act as a bridge between algebraic and trigonometric forms.

Comparing Coefficients

The comparison works by relating the linear coefficients to the coefficients of trigonometric functions in an equation. From our exercise, the coefficients of x and y in the linear equation can be directly compared to the trigonometric coefficients in its transformed version. This lets students visualize and understand that every linear relationship between two variables can also be expressed using trigonometry when angles come into the picture. It's a fabulous way to see the interconnectedness of mathematical concepts!

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any value of the variable within their domain. Examples include sin^2(θ) + cos^2(θ) = 1 and tan(θ) = sin(θ)/cos(θ). Identities allow us to express one trigonometric function in terms of others and are essential when solving trigonometric equations.

When working with inverse trigonometric functions to solve for an angle, it is crucial to remember the constraints of the principal values. The sin^-1 and cos^-1 functions return values within certain intervals, typically [-π/2, π/2] for sin^-1 and [0, π] for cos^-1, which is crucial for finding the accurate value of an angle that aligns with the given conditions, as shown in the exercise.