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

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Evaluate Composite Trig Functions

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Evaluate Composite Trig Functions: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Understanding composite trigonometric functions involves evaluating them from the inside out, starting with the inner function. For example, finding the sine of the inverse cosine of 1/2 leads to determining the angle first, then applying the outer function. Drawing right triangles can aid in evaluating functions like the sine of the inverse tangent of 3/4, where the sides represent the tangent ratio. Always consider the defined intervals for inverse functions to ensure accurate results, especially when dealing with negative values or undefined expressions.

concept

Evaluate Composite Functions - Values on Unit Circle

Video duration:

Evaluate Composite Functions - Values on Unit Circle Video Summary

Understanding composite trigonometric functions is essential for evaluating expressions involving inverse trigonometric functions. When tasked with finding the sine of the inverse cosine of $ \frac{1}{2} $, we can simplify the process by breaking it down into manageable parts. The first step is to evaluate the inside function, which in this case is the inverse cosine of $ \frac{1}{2} $. This can be interpreted as determining the angle whose cosine value is $ \frac{1}{2} $. Referring to the unit circle, we find that the angle corresponding to this cosine value is $ \frac{\pi}{3} $.

Next, we proceed to the outside function, which is the sine of $ \frac{\pi}{3} $. Again, using the unit circle, we see that the sine of $ \frac{\pi}{3} $ is $ \frac{\sqrt{3}}{2} $. Therefore, the overall expression evaluates to:

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

In another example, we consider the cosine of the inverse tangent of $ 0 $. Here, we first evaluate the inside function, the inverse tangent of $ 0 $, which asks for the angle whose tangent is $ 0 $. The angle $ 0 $ satisfies this condition, and since it falls within the defined interval for the inverse tangent (from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $), we can proceed. The cosine of $ 0 $ is $ 1 $, leading to the conclusion that:

\[\cos(\tan^{-1}(0)) = 1\]

Lastly, we explore the inverse cosine of the sine of $ \frac{\pi}{3} $. We start with the inside function, which is the sine of $ \frac{\pi}{3} $. From the unit circle, we know that this sine value is $ \frac{\sqrt{3}}{2} $. Now, we need to find the angle whose cosine is $ \frac{\sqrt{3}}{2} $. The relevant angle within the interval for the inverse cosine (from $ 0 $ to $ \pi $) is $ \frac{\pi}{6} $. Thus, we conclude that:

\[\cos^{-1}(\sin(\frac{\pi}{3})) = \frac{\pi}{6}\]

By systematically evaluating composite trigonometric functions, we can simplify complex expressions into straightforward calculations, enhancing our understanding of trigonometric relationships and their applications.

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

Video duration:

Example 1 Video Summary

To evaluate the expression for the secant of the inverse cosine of $-\frac{\sqrt{3}}{2}$, we start by determining the angle whose cosine equals $-\frac{\sqrt{3}}{2}$. This requires us to consider the range of the inverse cosine function, which is limited to angles between $0$ and $\pi$.

From the unit circle, we find that the angle $ \frac{5\pi}{6} $ has a cosine value of $-\frac{\sqrt{3}}{2}$. Thus, we can express the original problem as finding the secant of $ \frac{5\pi}{6} $.

Recall that the secant function is defined as the reciprocal of the cosine function. Therefore, we can rewrite the secant of $ \frac{5\pi}{6} $ as:

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

Since we already established that $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, we substitute this value into our equation:

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

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:

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

This concludes the evaluation, confirming that the secant of the inverse cosine of $-\frac{\sqrt{3}}{2}$ is $-\frac{2\sqrt{3}}{3}$.

concept

Evaluate Composite Functions - Special Cases

Video duration:

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. A common mistake is to assume that the inverse function cancels out the original function directly. For example, consider the expression $\cos^{-1}(\cos(\frac{11\pi}{6}))$. While it may seem intuitive to simplify this to $\frac{11\pi}{6}$, this is incorrect due to the defined interval of the inverse cosine function.

To solve $\cos^{-1}(\cos(\frac{11\pi}{6}))$, we first evaluate the inner function, which is $\cos(\frac{11\pi}{6})$. Referring to the unit circle, we find that $\cos(\frac{11\pi}{6}) = \frac{\sqrt{3}}{2}$. Next, we need to evaluate the outer function, $\cos^{-1}(\frac{\sqrt{3}}{2})$. The inverse cosine function is defined for angles between $0$ and $\pi$. Within this interval, the angle whose cosine is $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$. Thus, the final answer is $\cos^{-1}(\cos(\frac{11\pi}{6})) = \frac{\pi}{6}$.

In another example, consider $\sin(\sin^{-1}(2))$. Here, we start with the inner function $\sin^{-1}(2)$. The inverse sine function is only defined for values between $-1$ and $1$. Since $2$ is outside this range, $\sin^{-1}(2)$ is undefined. Therefore, attempting to evaluate $\sin(\sin^{-1}(2))$ leads to an error, reinforcing the importance of considering the defined intervals of inverse trigonometric functions.

In summary, when dealing with composite trigonometric functions, always evaluate from the inside out and be mindful of the defined intervals for inverse functions to avoid incorrect simplifications.

example

Example 2

Video duration:

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 range of 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

Video duration:

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 asks for 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 of 5, derived from the Pythagorean theorem. Thus, the sine of this angle is calculated as:

\[\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}\]

Next, consider evaluating the sine of the inverse cosine of $ -\frac{5}{13} $. Here, we first determine the appropriate quadrant for our angle. Since the inverse cosine function yields angles between $ 0 $ and $ \pi $, and the cosine value is negative, the angle must be in the second quadrant. We then draw a right triangle in this quadrant, labeling the adjacent side as $ -5 $ and the hypotenuse as 13. To find the opposite side, we again apply the Pythagorean theorem:

\[a^2 + (-5)^2 = 13^2\]

Solving this gives:

\[a^2 + 25 = 169 \implies a^2 = 144 \implies a = 12\]

Now, we can find the sine of the angle $ \theta $ using the opposite side and the hypotenuse:

\[\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13}\]

This result is consistent with the properties of the sine function in the second quadrant, where sine values are positive. By following these steps—drawing a right triangle, identifying 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

Video duration:

Example 3 Video Summary

To evaluate the expression for the sine of the inverse tangent of $ \frac{2}{3} $, we start by recognizing that the inverse tangent function, denoted as $ \tan^{-1} $, yields an angle whose tangent is $ \frac{2}{3} $. Since the range of the inverse tangent function is from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $, the angle will be located in the first quadrant where both sine and tangent are positive.

Next, we construct a right triangle to represent this situation. In this triangle, we label the angle $ \theta $ such that:

The opposite side (to angle $ \theta $) is $ 2 $.
The adjacent side is $ 3 $.

Using the Pythagorean theorem, we can find the hypotenuse $ c $ of the triangle. The theorem states:

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

Substituting the known values:

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

Calculating gives:

4 + 9 = c^2

Thus, $ c^2 = 13 $, leading to $ c = \sqrt{13} $.

Now, we can find the sine of angle $ \theta $. According to the definition of sine in a right triangle:

$ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $

Substituting the values 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} $

Therefore, the final result is that the sine of the inverse tangent of $ \frac{2}{3} $ is:

$ \sin(\tan^{-1}(\frac{2}{3})) = \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

Video duration:

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 SOHCAHTOA mnemonic, this tells us that the opposite side of the triangle is $ x $ and the hypotenuse is $ \sqrt{x^2 + 4} $.

To find the length of the adjacent side, we apply the Pythagorean theorem, which states that $ a^2 + b^2 = c^2 $. Here, we let $ a $ be the length of the adjacent side, $ b $ be $ x $, and $ c $ be $ \sqrt{x^2 + 4} $. Setting up the equation:

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

This simplifies to:

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

By subtracting $ x^2 $ from both sides, 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 that the tangent of the inverse sine of $ \frac{x}{\sqrt{x^2 + 4}} $ is equal to $ \frac{x}{2} $.

Do you want more practice?

We have more practice problems on Evaluate Composite Trig Functions

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 inner function. The inverse cosine of 1/2 asks for the angle whose cosine is 1/2. From the unit circle, this angle is π/3. Next, evaluate the outer function: the sine of π/3. From the unit circle, the sine of π/3 is √3/2. Therefore, the sine of the inverse cosine of 1/2 is √3/2.

What is the cosine of the inverse tangent of 0?

First, find the inverse tangent of 0, which asks for the angle whose tangent is 0. From the unit circle, this angle is 0. Next, evaluate the outer function: the cosine of 0. From the unit circle, the cosine of 0 is 1. Therefore, the cosine of the inverse tangent of 0 is 1.

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

Start with the inner function: the sine of π/3. From the unit circle, the sine of π/3 is √3/2. Next, find the inverse cosine of √3/2, which asks for the angle whose cosine is √3/2. From the unit circle, this angle is π/6. Therefore, the inverse cosine of the sine of π/3 is π/6.

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

First, find the inverse tangent of 3/4, which asks for the angle whose tangent is 3/4. Draw a right triangle with opposite side 3 and adjacent side 4. Use the Pythagorean theorem to find the hypotenuse: √(3² + 4²) = 5. Next, evaluate the outer function: the sine of this angle. The sine is the opposite side over the hypotenuse, which is 3/5. Therefore, the sine of the inverse tangent of 3/4 is 3/5.

What is the inverse cosine of the cosine of 11π/6?

First, find the cosine of 11π/6. From the unit circle, the cosine of 11π/6 is √3/2. Next, find the inverse cosine of √3/2, which asks for the angle whose cosine is √3/2 within the interval [0, π]. From the unit circle, this angle is π/6. Therefore, the inverse cosine of the cosine of 11π/6 is π/6.

How do you evaluate the sine of the inverse cosine of -5/13?

First, find the inverse cosine of -5/13, which asks for the angle whose cosine is -5/13. Draw a right triangle in quadrant II with adjacent side -5 and hypotenuse 13. Use the Pythagorean theorem to find the opposite side: √(13² - 5²) = 12. Next, evaluate the outer function: the sine of this angle. The sine is the opposite side over the hypotenuse, which is 12/13. Therefore, the sine of the inverse cosine of -5/13 is 12/13.