Problem 32 - Calculus with Concepts in Calculus

Question

a. Using \((9)\), show that \(\cos x=\sqrt{1-\sin ^{2} x}\) for \(0 \leq x \leq\) \(\pi / 2\) and for \(3 \pi / 2 \leq x \leq 2 \pi .\) Show also that \(\cos x=-\sqrt{1-\sin ^{2} x}\) for \(\pi / 2 \leq x \leq 3 \pi / 2\) b. Using \((9)\), show that \(\sin x=\sqrt{1-\cos ^{2} x}\) for \(0 \leq x \leq \pi\) Show also that \(\sin x=-\sqrt{1-\cos ^{2} x}\) for \(\pi \leq x \leq 2 \pi\)

Step-by-Step Solution

Answer

For part (a), use the identity and interval conditions to determine the sign of the square root. For part (b), apply similar logic to determine the sign of the sine expression. Apply these trigonometric identities with correct signs based on intervals.

Step 1: Understanding Trigonometric Identity

Recall the Pythagorean identity for trigonometric functions: \[ \cos^{2} x + \sin^{2} x = 1 \\] This identity holds for all values of \(x\). We will use this identity to solve the given problem.

Step 2: Derive Cosine Expression for Given Intervals (a)

By rearranging the identity, we get:\[ \cos^{2} x = 1 - \sin^{2} x \\]To solve for \(\cos x\), take the square root of both sides:\[\cos x = \pm \sqrt{1 - \sin^{2} x}\]For the intervals \(0 \leq x \leq \pi/2\) and \(3\pi/2 \leq x \leq 2\pi\), \(\cos x\) is non-negative. Thus, \(\cos x = \sqrt{1 - \sin^{2} x}\) for these intervals.

Step 3: Derive Cosine Expression for \\\pi/2 \leq x \leq 3\pi/2\\ Interval (a)

In the interval where \(\pi/2 \leq x \leq 3\pi/2\), \(\cos x\) is non-positive. Therefore, we use the negative root:\[ \cos x = -\sqrt{1 - \sin^{2} x} \] as \(\cos x\) is negative in this interval.

Step 4: Derive Sine Expression for Given Intervals (b)

Using the same identity, rearrange to find \(\sin^{2} x\): \[ \sin^{2} x = 1 - \cos^{2} x \]Take the square root of both sides to find \(\sin x\):\[ \sin x = \pm \sqrt{1 - \cos^{2} x} \]For \(0 \leq x \leq \pi\), \(\sin x\) is non-negative. Hence, \(\sin x = \sqrt{1 - \cos^{2} x}\) within this interval.

Step 5: Derive Sine Expression for \\\pi \leq x \leq 2\pi\\ Interval (b)

In the interval \(\pi \leq x \leq 2\pi\), sine is non-positive. Thus, \(\sin x = -\sqrt{1 - \cos^{2} x}\) applies here because \(\sin x\) is negative in this portion of the unit circle.

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry that helps us relate the sine and cosine of an angle. The identity is expressed as \( \cos^{2} x + \sin^{2} x = 1 \), which holds true for any angle \( x \). This identity is derived from the Pythagorean theorem and is named accordingly. It is similar to the equation \( a^2 + b^2 = c^2 \) for a right triangle, where \( a \) and \( b \) are the legs and \( c \) the hypotenuse.

• It enables us to find one trigonometric function given the other.
• Rearranging terms can yield expressions for both \( \cos x \) and \( \sin x \).

Understanding the Pythagorean identity is crucial as it is used to transform, simplify, and solve trigonometric equations.

Sine and Cosine

Trigonometric functions express relationships between the sides of a right triangle and its angles.
Sine (\( \sin \)) and cosine (\( \cos \)) are the most fundamental among them. They are defined as:

• \( \sin x \) is the ratio of the length of the opposite side to the hypotenuse.
• \( \cos x \) is the ratio of the length of the adjacent side to the hypotenuse.

These functions vary between -1 and 1 as angles change. The expressions \( \cos x = \pm \sqrt{1 - \sin^{2} x} \) and \( \sin x = \pm \sqrt{1 - \cos^{2} x} \) are derived from the Pythagorean identity.
Depending on the angle's quadrant, you select either the positive or negative sign:

• In quadrants where sine or cosine is positive, use the positive root.
• Negative roots apply where these functions are negative.

Knowing which sign to use is key in solving trigonometric problems and analyzing function behaviors.

Interval Analysis

Interval analysis helps determine the correct sign for sine and cosine expressions across different angle ranges. In trigonometry, the unit circle is used to evaluate angles and their corresponding sine and cosine values within specific intervals.
It's essential to understand how sine and cosine behave in each quadrant:

• In the first quadrant \( (0 \leq x \leq \pi/2) \), both \( \sin x \) and \( \cos x \) are positive.
• In the second quadrant \( (\pi/2 \leq x \leq \pi) \), \( \sin x \) is positive, while \( \cos x \) is negative.
• In the third quadrant \( (\pi \leq x \leq 3\pi/2) \), both are negative.
• In the fourth quadrant \( (3\pi/2 \leq x \leq 2\pi) \), \( \sin x \) is negative, and \( \cos x \) is positive.

By understanding interval analysis, you can accurately apply the correct signs for trigonometric expressions in various scenarios. This knowledge helps solve exercises like the one given, ensuring precise application of derived trigonometric identities for different angle ranges.