Evaluate the expression. sin ⁡ − 1 ( cos ⁡ 2 π 3 ) \sin^{-1}\left(\cos\frac{2\pi}{3}\right) sin − 1 ( cos 3 2 π )

− π 6 -\frac{\pi}{6} − 6 π

Evaluate the expression. tan ⁡ − 1 ( tan ⁡ 2 π 3 ) \tan^{-1}\left(\tan\frac{2\pi}{3}\right) tan − 1 ( tan 3 2 π )

− π 3 -\frac{\pi}{3} − 3 π

Evaluate the expression. cos ⁡ ( sin ⁡ − 1 ( − 7 25 ) ) \cos\left(\sin^{-1}\left(-\frac{7}{25}\right)\right) cos ( sin − 1 ( − 25 7 ) )

− 24 25 -\frac{24}{25} − 25 24

11. Inverse Trigonometric Functions and Basic Trig Equations

Evaluate Composite Trig Functions

11. Inverse Trigonometric Functions and Basic Trig Equations

Evaluate Composite Trig Functions: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

To evaluate composite trigonometric functions, start with the inside function, considering the appropriate intervals. For example, the sine of the inverse tangent of 3/4 can be solved by drawing a right triangle, leading to a sine value of 3/5. Similarly, for the inverse cosine of negative 5/13, identify the quadrant, draw the triangle, and apply the Pythagorean theorem to find the sine, resulting in 12/13. Understanding these principles is crucial for solving complex trigonometric expressions effectively.

concept

Evaluate Composite Functions - Values on Unit Circle

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Evaluate Composite Functions - Values on Unit Circle Video Summary

Understanding composite trigonometric functions is essential for evaluating expressions that involve multiple trigonometric operations. When faced with an expression like the sine of the inverse cosine of $ \frac{1}{2} $, it is helpful to break it down into its components. This expression can be viewed as a composite function, where we first evaluate the inner function, which in this case is the inverse cosine of $ \frac{1}{2} $.

To find the inverse cosine of $ \frac{1}{2} $, we ask ourselves: "What angle has a cosine value of $ \frac{1}{2} $?" Referring to the unit circle, we find that the angle $ \frac{\pi}{3} $ satisfies this condition. Thus, the inner function evaluates to $ \frac{\pi}{3} $. Next, we need to find the sine of this angle. The sine of $ \frac{\pi}{3} $ is known to be $ \frac{\sqrt{3}}{2} $. Therefore, the overall expression, the sine of the inverse cosine of $ \frac{1}{2} $, simplifies to $ \frac{\sqrt{3}}{2} $.

In another example, consider the cosine of the inverse tangent of $ 0 $. Again, we start with the inner function, which is the inverse tangent of $ 0 $. We determine which angle has a tangent value of $ 0 $. The angles $ 0 $ and $ \pi $ both yield a tangent of $ 0 $, but since the range of the inverse tangent function is limited to $ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $, we select $ 0 $ as our solution. Now, we find the cosine of $ 0 $, which is $ 1 $. Thus, the final result for this expression is $ 1 $.

Lastly, we can evaluate the inverse cosine of the sine of $ \frac{\pi}{3} $. Here, we first find the sine of $ \frac{\pi}{3} $, which we already know is $ \frac{\sqrt{3}}{2} $. Now, we need to determine the angle whose cosine is $ \frac{\sqrt{3}}{2} $. The angle $ \frac{\pi}{6} $ fits this criterion and lies within the range of the inverse cosine function, which is $ [0, \pi] $. Therefore, the inverse cosine of the sine of $ \frac{\pi}{3} $ evaluates to $ \frac{\pi}{6} $.

In summary, evaluating composite trigonometric functions involves systematically breaking down the expression, starting with the inner function and working outward. This method not only simplifies the process but also reinforces the understanding of the relationships between different trigonometric functions and their inverses.

Problem

Evaluate the expression.
$\cos\left(\sin^{-1}1\right)$

$0$

$1$

$-1$

$\frac12$

Problem

Evaluate the expression.
$\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$

$\frac{\pi}{6}$

$\frac{5\pi}{6}$

$\frac{\pi}{3}$

$-\frac{\pi}{6}$

example

Example 1

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Example 1 Video Summary

To evaluate the expression for the secant of the inverse cosine of $-\frac{\sqrt{3}}{2}$, we start by breaking down the components of the expression. The first step involves determining the angle whose cosine equals $-\frac{\sqrt{3}}{2}$. Since we are dealing with the inverse cosine function, we need to find an angle within the interval $[0, \pi]$.

From the unit circle, we know that the cosine of $\frac{5\pi}{6}$ is $-\frac{\sqrt{3}}{2}$. Therefore, we can conclude that:

\[\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}\]

Next, we need to find the secant of this angle. The secant function is defined as the reciprocal of the cosine function:

\[\sec\left(\frac{5\pi}{6}\right) = \frac{1}{\cos\left(\frac{5\pi}{6}\right)}\]

We already established that $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$. Substituting this value into the secant function gives us:

\[\sec\left(\frac{5\pi}{6}\right) = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}\]

To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

\[-\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\]

Thus, the final result for the secant of the inverse cosine of $-\frac{\sqrt{3}}{2}$ is:

\[\sec\left(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = -\frac{2\sqrt{3}}{3}\]

This process illustrates the relationship between inverse trigonometric functions and their corresponding angles, as well as the importance of rationalizing denominators in trigonometric expressions.

concept

Evaluate Composite Functions - Special Cases

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Evaluate Composite Functions - Special Cases Video Summary

When working with composite trigonometric functions, it's essential to understand the relationship between trigonometric functions and their inverses. For example, consider the expression inverse cosine of cosine of $ \frac{11\pi}{6} $. While it may seem intuitive to cancel the inverse cosine and cosine, resulting in $ \frac{11\pi}{6} $, this approach is incorrect due to the defined intervals of inverse trigonometric functions.

To solve such problems accurately, evaluate the inner function first. In this case, calculate the cosine of $ \frac{11\pi}{6} $. Referring to the unit circle, we find that $ \cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} $. Next, we evaluate the outer function, which is the inverse cosine of $ \frac{\sqrt{3}}{2} $. The question now becomes: what angle has a cosine value of $ \frac{\sqrt{3}}{2} $ within the interval $ [0, \pi] $? The answer is $ \frac{\pi}{6} $, confirming that the final answer is indeed $ \frac{\pi}{6} $, not $ \frac{11\pi}{6} $.

In another example, consider the expression sine of inverse sine of 2. Again, do not cancel the sine and inverse sine functions. Instead, evaluate the inner function first: what angle has a sine value of 2? However, the sine function only outputs values in the range of $[-1, 1]$. Since 2 is outside this interval, the expression is undefined. Attempting to compute this on a calculator would yield an error, reinforcing the importance of considering the defined intervals of inverse trigonometric functions.

In summary, always evaluate the inner function first and be mindful of the intervals for inverse trigonometric functions to avoid incorrect assumptions and errors in calculations.

example

Example 2

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Example 2 Video Summary

In evaluating the expression for the inverse sine of the sine of $\frac{\pi}{6}$, it is crucial to understand the behavior of inverse trigonometric functions and their defined intervals. The sine function, $\sin$, takes an angle and returns a ratio, while the inverse sine function, $\sin^{-1}$, returns an angle based on a given sine value, but only within a specific range.

First, we calculate the sine of $\frac{\pi}{6}$. Referring to the unit circle, we find that:

\[\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\]

Next, we need to find the inverse sine of $\frac{1}{2}$. This can be interpreted as determining the angle whose sine is $\frac{1}{2}$. However, we must ensure that this angle lies within the defined interval for the inverse sine function, which is from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.

Within this interval, the angle that satisfies $\sin(x) = \frac{1}{2}$ is:

\[x = \frac{\pi}{6}\]

Thus, we conclude that:

\[\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\]

It is important to note that while it may seem intuitive to cancel the sine and inverse sine functions, this is only valid when the original angle is within the specified interval of the inverse function. In this case, since $\frac{\pi}{6}$ is indeed within the interval $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, we can safely conclude that:

\[\sin^{-1}(\sin(\frac{\pi}{6})) = \frac{\pi}{6}\]

Always remember to verify the interval when working with composite trigonometric functions to ensure accurate results.

Problem

Evaluate the expression.
$\cos\left(\cos^{-1}\left(-\sqrt3\right)\right)$

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$\pi$

Undefined

Problem

Evaluate the expression.
$\cos^{-1}\left(\cos\left(\frac{\pi}{2}\right)\right)$

$0$

$\frac{\pi}{2}$

$\pi$

Undefined

Problem

Evaluate the expression.
$\tan^{-1}\left(\tan\frac{2\pi}{3}\right)$

$-\frac{\pi}{3}$

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$\frac{5\pi}{3}$

concept

Evaluate Composite Functions - Values Not on Unit Circle

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Evaluate Composite Functions - Values Not on Unit Circle Video Summary

Evaluating composite trigonometric functions can be simplified by using right triangles, especially when the values are not directly found on the unit circle. For instance, to evaluate the sine of the inverse tangent of $ \frac{3}{4} $, we start by recognizing that the inverse tangent function gives us an angle $ \theta $ such that $ \tan(\theta) = \frac{3}{4} $. This can be visualized by drawing a right triangle where the opposite side is 3 and the adjacent side is 4, leading to a hypotenuse calculated using the Pythagorean theorem:

$ a^2 + b^2 = c^2 $

$ 3^2 + 4^2 = c^2 $

$ 9 + 16 = c^2 $

$ c^2 = 25 $

$ c = 5 $

With the hypotenuse determined, we can find the sine of the angle $ \theta $ using the sine definition from SOHCAHTOA, which states that $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $. Thus, $ \sin(\theta) = \frac{3}{5} $. Therefore, the sine of the inverse tangent of $ \frac{3}{4} $ is $ \frac{3}{5} $.

In another example, to evaluate the sine of the inverse cosine of $ -\frac{5}{13} $, we again start with the inside function. The inverse cosine function gives us an angle $ \theta $ such that $ \cos(\theta) = -\frac{5}{13} $. Since cosine is negative, we know that $ \theta $ must be in the second quadrant. We draw a right triangle in this quadrant, labeling the adjacent side as $ -5 $ and the hypotenuse as $ 13 $. To find the opposite side, we apply the Pythagorean theorem:

$ a^2 + b^2 = c^2 $

$ (-5)^2 + b^2 = 13^2 $

$ 25 + b^2 = 169 $

$ b^2 = 144 $

$ b = 12 $

Now, we can find the sine of angle $ \theta $ using SOHCAHTOA, where $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $. Thus, $ \sin(\theta) = \frac{12}{13} $. This means that the sine of the inverse cosine of $ -\frac{5}{13} $ is $ \frac{12}{13} $. This result is consistent with the fact that sine values are positive in the second quadrant.

By following these steps—drawing a right triangle, labeling the sides based on the trigonometric function, and applying the Pythagorean theorem—you can effectively evaluate any composite trigonometric function, regardless of whether the values are on the unit circle.

example

Example 3

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Example 3 Video Summary

To evaluate the expression for the sine of the inverse tangent of $ \frac{2}{3} $, we can utilize a right triangle approach since the angle corresponding to the inverse tangent is not a standard angle on the unit circle. The process involves several steps to derive the final result.

First, we identify the quadrant for the angle derived from the inverse tangent function. The range of the inverse tangent function is from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $, which indicates that the angle will be in either the first or second quadrant. Given that the argument $ \frac{2}{3} $ is positive, we conclude that the angle must be in the first quadrant.

Next, we draw a right triangle in the first quadrant. We label the angle $ \theta $ and use the definition of tangent, which is the ratio of the opposite side to the adjacent side. Since $ \tan(\theta) = \frac{2}{3} $, we assign the opposite side a length of 2 and the adjacent side a length of 3.

To find the hypotenuse, we apply the Pythagorean theorem, which states that $ a^2 + b^2 = c^2 $. Here, $ a = 2 $ and $ b = 3 $. Thus, we calculate:

$ 2^2 + 3^2 = c^2 $

$ 4 + 9 = c^2 $

$ 13 = c^2 $

Taking the square root of both sides gives us the hypotenuse $ c = \sqrt{13} $.

Now, we can evaluate the sine of the angle $ \theta $. According to the sine definition, $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $. Therefore, we have:

$ \sin(\theta) = \frac{2}{\sqrt{13}} $

To express this in a more standard form, we rationalize the denominator by multiplying the numerator and denominator by $ \sqrt{13} $:

$ \sin(\theta) = \frac{2 \sqrt{13}}{13} $

Thus, the final result is that the sine of the inverse tangent of $ \frac{2}{3} $ is equal to $ \frac{2 \sqrt{13}}{13} $.

Problem

Evaluate the expression.
$\tan\left(\cos^{-1}\frac{12}{13}\right)$

$\frac{12}{13}$

$\frac{12}{5}$

$\frac{5}{13}$

$\frac{5}{12}$

Problem

Evaluate the expression.
$\sin\left(\tan^{-1}\frac{15}{8}\right)$

$\frac{8}{15}$

$\frac{15}{17}$

$\frac{8}{17}$

$\frac{15}{289}$

Problem

Evaluate the expression.
$\cos\left(\sin^{-1}\left(-\frac{7}{25}\right)\right)$

$\frac{7}{24}$

$\frac{8}{25}$

$\frac{24}{25}$

$-\frac{24}{25}$

example

Example 4

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Example 4 Video Summary

To evaluate the expression for the tangent of the inverse sine of $ \frac{x}{\sqrt{x^2 + 4}} $, we can break it down into manageable steps. First, we need to understand the range of the inverse sine function, which is defined for angles between $ -\frac{\pi}{2} $ and $ \frac{\pi}{2} $. This means that the sine values are positive in the first quadrant, where our triangle will be located.

Next, we can construct a right triangle to represent the angle $ \theta $ such that $ \sin(\theta) = \frac{x}{\sqrt{x^2 + 4}} $. According to the sine definition (SOHCAHTOA), the opposite side corresponds to $ x $ and the hypotenuse corresponds to $ \sqrt{x^2 + 4} $. To find the adjacent side, we apply the Pythagorean theorem, which states that $ a^2 + b^2 = c^2 $. Here, we let $ a $ be the adjacent side, $ b $ be $ x $, and $ c $ be $ \sqrt{x^2 + 4} $.

Setting up the equation, we have:

$ a^2 + x^2 = (\sqrt{x^2 + 4})^2 $

which simplifies to:

$ a^2 + x^2 = x^2 + 4 $

By isolating $ a^2 $, we find:

$ a^2 = 4 $

Taking the square root gives us $ a = 2 $. Now we have all the sides of our triangle: the opposite side is $ x $, the adjacent side is $ 2 $, and the hypotenuse is $ \sqrt{x^2 + 4} $.

Finally, we can evaluate the tangent of $ \theta $. The tangent function is defined as the ratio of the opposite side to the adjacent side:

$ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{2} $

Thus, the final result is:

$ \tan(\sin^{-1}(\frac{x}{\sqrt{x^2 + 4}})) = \frac{x}{2} $

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Here’s what students ask on this topic:

How do you evaluate the sine of the inverse cosine of 1/2?

To evaluate the sine of the inverse cosine of 1/2, start with the inside function, which is the inverse cosine of 1/2. This asks for the angle whose cosine is 1/2. From the unit circle, we know that cos(π/3) = 1/2. So, the inverse cosine of 1/2 is π/3. Next, evaluate the sine of π/3. From the unit circle, sin(π/3) = √3/2. Therefore, the sine of the inverse cosine of 1/2 is √3/2.

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What is the cosine of the inverse tangent of 0?

To find the cosine of the inverse tangent of 0, start with the inside function, which is the inverse tangent of 0. This asks for the angle whose tangent is 0. From the unit circle, tan(0) = 0. So, the inverse tangent of 0 is 0. Next, evaluate the cosine of 0. From the unit circle, cos(0) = 1. Therefore, the cosine of the inverse tangent of 0 is 1.

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How do you solve the inverse cosine of the sine of π/3?

To solve the inverse cosine of the sine of π/3, start with the inside function, which is the sine of π/3. From the unit circle, sin(π/3) = √3/2. Next, evaluate the inverse cosine of √3/2. This asks for the angle whose cosine is √3/2. From the unit circle, cos(π/6) = √3/2. Therefore, the inverse cosine of the sine of π/3 is π/6.

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How do you evaluate the sine of the inverse tangent of 3/4?

To evaluate the sine of the inverse tangent of 3/4, start with the inside function, which is the inverse tangent of 3/4. This asks for the angle whose tangent is 3/4. Draw a right triangle where the opposite side is 3 and the adjacent side is 4. Using the Pythagorean theorem, the hypotenuse is √(3² + 4²) = 5. Now, evaluate the sine of this angle, which is the opposite side over the hypotenuse: sin(θ) = 3/5. Therefore, the sine of the inverse tangent of 3/4 is 3/5.

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What is the sine of the inverse cosine of -5/13?

To find the sine of the inverse cosine of -5/13, start with the inside function, which is the inverse cosine of -5/13. This asks for the angle whose cosine is -5/13. Draw a right triangle in the second quadrant where the adjacent side is -5 and the hypotenuse is 13. Using the Pythagorean theorem, the opposite side is √(13² - 5²) = 12. Now, evaluate the sine of this angle, which is the opposite side over the hypotenuse: sin(θ) = 12/13. Therefore, the sine of the inverse cosine of -5/13 is 12/13.

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