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Chapter 0: Problem 53

If \(\sin t=0.8\) and \(\cos t<0,\) find each value. (a) \(\sin (-t)\) (b) \(\cos t\) (c) \(\sin 2 t\) (d) \(\tan t\) (e) \(\cos \left(\frac{\pi}{2}-t\right)\) (f) \(\sin (\pi+t)\)

Short Answer

Expert verified
(a) -0.8; (b) -0.6; (c) -0.96; (d) -4/3; (e) 0.8; (f) -0.8.

Step by step solution

01

Determine the Quadrant

Given that \( \sin t = 0.8 \) and \( \cos t < 0 \), the angle \( t \) must be in the second quadrant. In this quadrant, sine is positive and cosine is negative.
02

Find \( \sin(-t) \)

By the identity \( \sin(-t) = -\sin t \), we know \( \sin(-t) = -0.8 \).
03

Find \( \cos t \) using Pythagorean Identity

Using the identity \( \sin^2 t + \cos^2 t = 1 \), substitute \( \sin t = 0.8 \) into the equation to find \( \cos t \). \( (0.8)^2 + \cos^2 t = 1 \), leading to \( 0.64 + \cos^2 t = 1 \). Hence, \( \cos^2 t = 0.36 \). As \( \cos t < 0 \), this means \( \cos t = -0.6 \).
04

Calculate \( \sin 2t \)

Using the double angle identity \( \sin 2t = 2 \sin t \cos t \), substitute \( \sin t = 0.8 \) and \( \cos t = -0.6 \). Thus, \( \sin 2t = 2 \times 0.8 \times (-0.6) = -0.96 \).
05

Find \( \tan t \)

The tangent is given by \( \tan t = \frac{\sin t}{\cos t} \). Hence, \( \tan t = \frac{0.8}{-0.6} = -\frac{4}{3} \).
06

Find \( \cos \left(\frac{\pi}{2}-t\right) \)

By the co-function identity, \( \cos\left(\frac{\pi}{2} - t\right) = \sin t \). Thus, \( \cos\left(\frac{\pi}{2} - t\right) = 0.8 \).
07

Calculate \( \sin (\pi + t) \)

By using the identity \( \sin(\pi + t) = -\sin t \), we get \( \sin(\pi + t) = -0.8 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadrant Identification
Quadrant identification is essential for understanding the signs of trigonometric functions. When given \( \sin t = 0.8 \) and \( \cos t < 0 \), the angle \( t \) can be identified to lie in the second quadrant.
In the coordinate plane, angles are measured in four quadrants:
  • The first quadrant (0 to \( \frac{\pi}{2} \)) has both \( \sin \) and \( \cos \) positive.
  • The second quadrant (\( \frac{\pi}{2} \) to \( \pi \)) has \( \sin \) positive, but \( \cos \) negative.
  • The third quadrant (\( \pi \) to \( \frac{3\pi}{2} \)) shows both \( \sin \) and \( \cos \) as negative.
  • The fourth quadrant (\( \frac{3\pi}{2} \) to \( 2\pi \)) keeps \( \sin \) negative, but \( \cos \) positive.
This identification helps to choose correct signs for trigonometric calculations. Since \( \sin t \) is positive and \( \cos t \) is negative, angle \( t \) resides in the second quadrant.
Pythagorean Identity
The Pythagorean Identity is a fundamental relationship in trigonometry, expressed as \( \sin^2 t + \cos^2 t = 1 \). This identity allows for finding unknown trigonometric function values when one function is known.
In our case, given \( \sin t = 0.8 \), substituting it into the formula helps find \( \cos t \):
  • First, calculate \( (0.8)^2 = 0.64 \).
  • Then the equation becomes \( 0.64 + \cos^2 t = 1 \).
  • Subtract 0.64 from 1 to get \( \cos^2 t = 0.36 \).
  • Taking the square root, \( \cos t = \pm 0.6 \).
Since \( \cos t < 0 \), \( \cos t \) must be \( -0.6 \).
This demonstrates how the Pythagorean Identity relates sine and cosine values for any angle.
Double Angle Identity
The Double Angle Identity helps find trigonometric values for double an angle. It is particularly useful in deriving angles that would be cumbersome otherwise. For sine, the identity is expressed as \( \sin 2t = 2 \sin t \cos t \).
In the problem, we use this identity with known values \( \sin t = 0.8 \) and \( \cos t = -0.6 \):
  • Substitute into the identity: \( \sin 2t = 2 \times 0.8 \times (-0.6) \).
  • Calculating gives \( \sin 2t = 2 \times 0.8 \times -0.6 = -0.96 \).
Understanding this identity helps in solving more complex trigonometric problems efficiently. It clearly shows how angles and their functions relate and transform.
Co-Function Identity
Co-function identities provide relationships between different trigonometric functions of complementary angles. For example, \( \cos(\frac{\pi}{2} - t) = \sin t \). This means the cosine of an angle's complement equals the sine of the angle.
In the exercise, given \( \sin t = 0.8 \), the identity gives:
  • \( \cos(\frac{\pi}{2} - t) = \sin t = 0.8 \).
Co-function identities bridge different trigonometric functions, often simplifying problems significantly if used correctly.

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Most popular questions from this chapter

Circular motion can be modeled by using the parametric representations of the form \(x(t)=\sin t\) and \(y(t)=\cos t\) (A parametric representation means that a variable, \(t\) in this case, determines both \(x(t)\) and \(y(t) .\) This will give the full circle for \(0 \leq t \leq 2 \pi .\) If we consider a 4 -foot-diameter wheel making one complete rotation clockwise once every 10 seconds, show that the motion of a point on the rim of the wheel can be represented by \(x(t)=2 \sin (\pi t / 5)\) and \(y(t)=2 \cos (\pi t / 5)\) (a) Find the positions of the point on the rim of the wheel when \(t=2\) seconds, 6 seconds, and 10 seconds. Where was this point when the wheel started to rotate at \(t=0 ?\) (b) How will the formulas giving the motion of the point change if the wheel is rotating counterclockwise. (c) At what value of \(t\) is the point at (2,0) for the first time?

The circular frequency \(v\) of oscillation of a point is given by \(v=\frac{2 \pi}{\text { period }}\). What happens when you add two motions that have the same frequency or period? To investigate, we can graph the functions \(y(t)=2 \sin (\pi t / 5)\) and \(y(t)=\sin (\pi t / 5)+\) \(\cos (\pi t / 5)\) and look for similarities. Armed with this information, we can investigate by graphing the following functions over the interval [-5,5] (a) \(y(t)=3 \sin (\pi t / 5)-5 \cos (\pi t / 5)+2 \sin ((\pi t / 5)-3)\) (b) \(y(t)=3 \cos (\pi t / 5-2)+\cos (\pi t / 5)+\cos ((\pi t / 5)-3)\)

Plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs (see Example 4). $$ \begin{array}{l} y=-2 x+3 \\ y=3 x^{2}-3 x+12 \end{array} $$

Plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs (see Example 4). $$ \begin{array}{l} y=-2 x+3 \\ y=-2(x-4)^{2} \end{array} $$

Write \(F(x)=\sqrt{1+\sin ^{2} x}\) as the composite of four functions, \(f \circ g \circ h \circ k\).
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