Question

The value of $\theta (0\le \theta \le \pi /2)$ satisfying the equation ${\sin }^2\theta -2\cos \theta +\frac{1}{4}=0$ is: (a) $\frac{\pi }{2}$ (b) $\frac{\pi }{3}$ (c) $\frac{\pi }{4}$ (d) $\frac{\pi }{6}$

Short answer

Answer: (b) $\frac{\pi }{3}$.

Step-by-step solution

Substitute Trigonometric Identity

Since we have ${\sin }^2\theta$ in our equation, we can replace it with $(1-{\cos }^2\theta )$ using the trigonometric identity ${\sin }^2\theta +{\cos }^2\theta =1$. The equation becomes: $$1-{\cos }^2\theta -2\cos \theta +\frac{1}{4}=0.$$

Simplify the Equation

Now we have a quadratic equation in terms of $\cos \theta$. Let's simplify it by multiplying both sides of the equation by 4 to eliminate the fraction: $$4-4{\cos }^2\theta -8\cos \theta +1=0.$$ This simplifies to: $$4{\cos }^2\theta +8\cos \theta -3=0.$$

Factor the Quadratic Equation

Now, we need to find the factors of the quadratic equation, $$4{\cos }^2\theta +8\cos \theta -3=(2\cos \theta -1)(2\cos \theta +3).$$

Solve for $\cos \theta$

We have two possible cases for $\cos \theta$: 1. $2\cos \theta -1=0\Rightarrow \cos \theta =\frac{1}{2}$. 2. $2\cos \theta +3=0\Rightarrow \cos \theta =-\frac{3}{2}$.

Evaluate the Possibilities

Now, let's evaluate the possibilities for $\theta$: 1. For $\cos \theta =\frac{1}{2}$, we have $\theta =\arccos \left(\frac{1}{2}\right)=\frac{\pi }{3}$. This value is within the given range $0\le \theta \le \frac{\pi }{2}$. 2. For $\cos \theta =-\frac{3}{2}$, since the maximum value of the cosine function is 1 and the minimum value is -1, it is not possible to have $\cos \theta =-\frac{3}{2}$. From the analysis, we conclude that the correct value of $\theta$ satisfying the given equation is $\theta =\frac{\pi }{3}$. Therefore, the answer is (b) $\frac{\pi }{3}$.

Key concepts

Trigonometric Identities

Trigonometric identities are equations that hold true for all values within specific domains of the trigonometric functions involved. They are pivotal in simplifying and solving trigonometric equations. A fundamental identity is ${\sin }^2\theta +{\cos }^2\theta =1$, which relates the square of the sine function and the square of the cosine function to 1. By understanding and applying this identity, students can transform trigonometric equations into more manageable forms, often reducing them to algebraic equations that are easier to solve.

In the given exercise, the identity helped to replace ${\sin }^2\theta$ with $1-{\cos }^2\theta$, effectively converting the trigonometric equation into a quadratic equation. This step is crucial for further simplification and allows for the application of algebraic methods to find the solution.

Quadratic Equations

A quadratic equation is an algebraic expression of the second degree, generally represented as $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants and $aeq0$. They have a characteristic 'U' shaped curve known as a parabola when plotted on a graph. Solving a quadratic equation entails finding the values of $x$ that make the equation true. There are multiple methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula.

During the solution of the exercise, after applying the trigonometric identity, the trigonometric equation was transformed into a quadratic equation with $\cos \theta$ as the variable. Solving this equation then became the focus to find the possible values of $\cos \theta$ that satisfy the original trigonometric equation.

Factors of Quadratic Equation

Factoring a quadratic equation involves breaking it down into simpler expressions, or factors, that multiply to give the original quadratic equation. The standard approach when $a=1$ is to find two numbers that multiply to $ac$ and add up to $b$, although more complex methods are required for other cases. Factoring can make solving quadratic equations straightforward, because it utilizes the zero product property which states that if the product of two factors is zero, then at least one of the factors must be zero.

In the exercise's solution, factoring was used to find the factors of the transformed quadratic equation. Once the factors were identified as $2\cos \theta -1$ and $2\cos \theta +3$, the zero product property was applied, which split the problem into two simpler equations to solve for $\cos \theta$. This process allowed for the subsequent determination of the angle $\theta$ values.

Inverse Trigonometric Functions

Inverse trigonometric functions allow calculating an angle from the known value of a trigonometric function. They are denoted with an 'arc' prefix, as in $\arcsin$, $\arccos$, and $\arctan$, or sometimes with a superscript '-1', as in ${\sin }^{-1}$, ${\cos }^{-1}$, and ${\tan }^{-1}$. It is important to note that these functions are not truly inverse functions but are properly called 'inverse trigonometric functions' due to certain limitations and conditions required to obtain unique values.

In the context of solving the provided trigonometric equation, once $\cos \theta$ was found to be $\frac{1}{2}$, the inverse trigonometric function $\arccos$ was employed to determine the angle $\theta$ that corresponds with that cosine value. Care was taken to ensure that the calculated angle lies within the specified range, thus leading to the correct solution.