Problem 51
Question
Show that the equation is not an Identity. $$\cos t=\sqrt{1-\sin ^{2} t}$$
Step-by-Step Solution
Verified Answer
The equation is not an identity since it fails for values like \( t = \pi \).
1Step 1: Understanding the Equation
The given equation is \( \cos t = \sqrt{1 - \sin^2 t} \). We need to determine if this is an identity, meaning it must hold for all possible values of \( t \).
2Step 2: Recalling the Pythagorean Identity
Recall the Pythagorean identity: \( \sin^2 t + \cos^2 t = 1 \). This identity is valid for all angles \( t \). By rearranging, we have \( \cos^2 t = 1 - \sin^2 t \).
3Step 3: Analyze the Given Equation
Compare the given equation \( \cos t = \sqrt{1 - \sin^2 t} \) with \( \cos^2 t = 1 - \sin^2 t \). From the identity, \( \cos^2 t = (\cos t)^2 \), hence, \( \cos t = \pm \sqrt{1 - \sin^2 t} \).
4Step 4: Evaluating Sign of Cosine
Since \( \cos t \) could be either \( \sqrt{1 - \sin^2 t} \) or \(-\sqrt{1 - \sin^2 t} \) depending on the quadrant that \( t \) is in, the equation \( \cos t = \sqrt{1 - \sin^2 t} \) is only true for angles where \( \cos t \geq 0 \).
5Step 5: Counterexample with a Specific Angle
To show the equation is not an identity, choose an example where \( \cos t < 0 \). For instance, let \( t = \pi \), then \( \cos(\pi) = -1 \) and \( \sqrt{1-\sin^2(\pi)} = \sqrt{1} = 1 \). Clearly, \(-1 eq 1\), showing the equation is not true for \( t = \pi \).
6Step 6: Conclusion
Since \( \cos t \) can be negative for certain values of \( t \), and the equation given always assumes a non-negative value, the equation \( \cos t = \sqrt{1 - \sin^2 t} \) is not an identity.
Key Concepts
Cosine FunctionPythagorean IdentityQuadrants of the Unit Circle
Cosine Function
The cosine function, often abbreviated as cos, is a fundamental trigonometric function. It is crucial in understanding angles and their properties in mathematics, especially within the context of a right-angled triangle. In simple terms, the cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.
In the unit circle, which is a circle with a radius of one centered at the origin of a coordinate plane, the cosine function becomes even more straightforward:
- The x-coordinate of a point on the unit circle is the cosine of the angle measured from the positive x-axis.
- As the angle increases, the cosine value represents the horizontal distance from the origin to the circle.
Pythagorean Identity
The Pythagorean Identity is a cornerstone of trigonometric equations, stemming from the Pythagorean Theorem. The identity can be stated as follows: \[sin^2 t + \cos^2 t = 1\]This identity is always true for any angle \( t \) and forms a foundation for many trigonometric transformations and proofs.By rearranging, we get another useful expression: \[\cos^2 t = 1 - \sin^2 t\]This derived equation shows that you can find \( \cos^2 t \) if you know \( \sin^2 t \), and vice versa. However, it's crucial to recognize that taking the square root of each side implies two possible roots:
- A positive root corresponding to \( \cos t = \sqrt{1 - \sin^2 t} \)
- A negative root corresponding to \( \cos t = -\sqrt{1 - \sin^2 t} \)
Quadrants of the Unit Circle
The unit circle is divided into four quadrants, each corresponding to a different range of angle measures. These quadrants are crucial for determining the sign of trigonometric functions like sine and cosine. Understanding the quadrant system can help you analyze angles and their corresponding trigonometric values accurately.
- First Quadrant: Angles from 0 to \( \frac{\pi}{2} \) (0 to 90 degrees). Here, both sine and cosine are positive.
- Second Quadrant: Angles from \( \frac{\pi}{2} \) to \( \pi \) (90 to 180 degrees). Sine is positive, and cosine is negative.
- Third Quadrant: Angles from \( \pi \) to \( \frac{3\pi}{2} \) (180 to 270 degrees). Both sine and cosine are negative.
- Fourth Quadrant: Angles from \( \frac{3\pi}{2} \) to \( 2\pi \) (270 to 360 degrees). Sine is negative, and cosine is positive.
Other exercises in this chapter
Problem 50
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$2 \sin ^{2} u+\sin u-6=0$$
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