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Chapter 4: Problem 63

Evaluate the following expressions exactly: $$\tan \left(-315^{\circ}\right)$$

Short Answer

Expert verified
\( \tan(-315^{\circ}) = 1 \).

Step by step solution

01

Convert Negative Angle to Positive

To find \( \tan(-315^{\circ}) \), first convert the negative angle to a positive angle. Add \( 360^{\circ} \) to \( -315^{\circ} \) to get \( -315^{\circ} + 360^{\circ} = 45^{\circ} \). This uses the periodicity of the tangent function, which has a period of \( 180^{\circ} \), to simplify the angle.
02

Evaluate Tangent of the Positive Angle

Now that we have converted the angle to \( 45^{\circ} \), evaluate \( \tan(45^{\circ}) \). The tangent of \( 45^{\circ} \) is known to be 1. Therefore, \( \tan(45^{\circ}) = 1 \).

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Key Concepts

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

Angle Conversion
Angle conversion is a fundamental concept when working with trigonometric functions. In trigonometry, angles can be measured in degrees or radians. Converting between negative and positive angles helps solve problems easily. When faced with a negative angle, we add a full rotation of \(360^{\circ}\) (or \(2\pi\) radians) to find the equivalent positive angle.
This process doesn't alter the trigonometric properties of the angle.
  • Convert \(-315^{\circ}\) to a positive angle by adding \(360^{\circ}\).
  • This results in the positive angle of \(45^{\circ}\).
This step ensures our calculations remain within the standard cycle of trigonometric functions, making it simpler to evaluate functions like tangent.
Periodicity of Tangent
The periodicity of the tangent function is a key property that simplifies complex angle evaluations. Unlike sine and cosine, with a period of \(360^{\circ}\) (or \(2\pi\) radians), tangent has a period of \(180^{\circ}\) (or \(\pi\) radians). This means the tangent function repeats its values every \(180^{\circ}\).
  • For instance, \(\tan(\theta) = \tan(\theta + 180^{\circ})\).
  • In our example, converting \(-315^{\circ}\) to \(45^{\circ}\) still results in the same tangent value due to this periodicity.
This property is particularly useful in converting angles and simplifying calculations, as different angles that differ by multiples of \(180^{\circ}\) will have identical tangent values.
Tangent of 45 Degrees
The tangent of \(45^{\circ}\) is a widely known and straightforward value in trigonometry.
The tangent function, defined as the ratio of the opposite side to the adjacent side in a right triangle, takes a special meaning at \(45^{\circ}\).
At this angle, both sides in a right-angled triangle are equal, making the tangent value 1.
Thus we have:
  • \(\tan(45^{\circ}) = \frac{1}{1} = 1\).
Remembering these simple angles and their tangent values helps tremendously when evaluating or verifying trigonometric expressions, providing a quick reference to frequently used results.

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Most popular questions from this chapter

Show that \(\frac{\cos \alpha}{a}+\frac{\cos \beta}{b}+\frac{\cos \gamma}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\) (Hint: Use the Law of Cosines.)

Find the distance a point travels along a circle over a time \(t,\) given the angular speed \(\omega\) and radius \(r\) of the circle. Round your answers to three significant digits. $$r=5 \mathrm{cm}, \omega=\frac{\pi \mathrm{rad}}{6 \mathrm{sec}}, t=10 \mathrm{sec}$$

Find the distance a point travels along a circle over a time \(t,\) given the angular speed \(\omega\) and radius \(r\) of the circle. Round your answers to three significant digits. $$r=2 \mathrm{mm}, \omega=6 \pi \frac{\mathrm{rad}}{\mathrm{sec}}, t=11 \mathrm{sec}$$

Determine whether each statement is true or false. All acute triangles can be solved using the Law of Cosines.

Explain the mistake that is made. If a bicycle has tires with radius 10 inches and the tires rotate \(90^{\circ}\) per \(\frac{1}{2}\) second, how fast is the bicycle traveling (linear speed) in miles per hour? Solution: Write the formula for linear speed. \(v=r \omega\) Let \(r=10\) inches and \(\omega=180^{\circ}\) per second. $$v=(10 \text { in. })\left(\frac{180^{\circ}}{\sec }\right)$$ Simplify. \( v=\frac{1800 \mathrm{in} .}{\mathrm{sec}}\) Let 1 mile \(=5280\) feet \(=63,360\) inches and 1 hour \(=3600\) seconds. $$v=\left(\frac{1800 \cdot 3600}{63,360}\right) \mathrm{mph}$$ Simplify. \(v \approx 102.3 \mathrm{mph}\) This is incorrect. The correct answer is approximately 1.8 mph. What mistake was made?
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