Learn how to solve trigonometric equations of the form sin(x) = 1, cos(x) = 1, and tan(x) = 1 over the interval [0, 2π). Explore the solutions graphically using the unit circle to understand the relationships between the sine, cosine, and tangent functions. Discover how these equations relate to the x and y-coordinates of points on the unit circle. Find step-by-step explanations and visual representations for each equation. Master the concepts of trigonometric functions and their solutions with clear and concise examples.
Trigonometric Equation
\( \cos(x) = 1 \)
Solutions within \( [0, 2\pi) \):
\( x = 0 \) and \( x = 2\pi \)
Explanation:
In the unit circle, the x-coordinate represents the cosine of an angle, and the y-coordinate represents the sine of the angle. When \( \cos(x) = 1 \), it means the x-coordinate of a point on the unit circle is 1.
At \( x = 0 \), the point on the unit circle lies at (1, 0), which is the rightmost point on the unit circle. This is the angle where the cosine function is maximum (1) and the sine function is 0.
Similarly, at \( x = 2\pi \), the point on the unit circle also lies at (1, 0), which is the same as the starting point. This is because the cosine function is periodic with a period of \( 2\pi \), and it repeats its maximum value at this angle.
Refer to the unit circle image above for a visual representation.
Trigonometric Equation
\( \sin(x) = 1 \)
Solutions within \( [0, 2\pi) \):
\( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \)
Explanation:
In the unit circle, the x-coordinate represents the cosine of an angle, and the y-coordinate represents the sine of the angle. When \( \sin(x) = 1 \), it means the y-coordinate of a point on the unit circle is 1.
At \( x = \frac{\pi}{2} \), the point on the unit circle lies at \(\left(0, 1\right)\), which is the topmost point on the unit circle. This is the angle where the sine function is maximum (1) and the cosine function is 0.
Similarly, at \( x = \frac{3\pi}{2} \), the point on the unit circle lies at \(\left(0, -1\right)\), which is the bottommost point on the unit circle. This is the angle where the sine function is minimum (-1) and the cosine function is 0.
Refer to the unit circle image above for a visual representation.
Trigonometric Equation
\( \tan(x) = 1 \)
Solutions within \( [0, 2\pi) \):
\( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \)
Explanation:
In the unit circle, the tangent of an angle \( x \) is defined as the ratio of the sine to the cosine at that angle, \( \tan(x) = \frac{\sin(x)}{\cos(x)} \).
At \( x = \frac{\pi}{4} \), the point on the unit circle lies at \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\). The tangent of \( x = \frac{\pi}{4} \) is \( \tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 \).
Similarly, at \( x = \frac{5\pi}{4} \), the point on the unit circle lies at \(\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\). The tangent of \( x = \frac{5\pi}{4} \) is \( \tan\left(\frac{5\pi}{4}\right) = \frac{\sin\left(\frac{5\pi}{4}\right)}{\cos\left(\frac{5\pi}{4}\right)} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 \).
Refer to the unit circle image above for a visual representation.