Finding the Exact Value of ( cos(arcsin(frac{8}{17})) ) Without a Calculator: A Step-by-Step Guide

Many calculus students and mathematicians encounter this problem: How can you find the exact value of ( cos(arcsin(frac{8}{17})) ) without using a calculator? This guide provides a detailed explanation using fundamental properties and the Pythagorean theorem.

Understanding the Problem

Let's define theta arcsin(frac{8}{17}). By definition, this means:

[ sin(theta) frac{8}{17} ]

We are tasked with finding the exact value of ( cos(theta) ).

Using the Pythagorean Identity

The Pythagorean identity for trigonometric functions is a powerful tool:

[ sin^2(theta) cos^2(theta) 1 ]

We can rearrange this to find ( cos^2(theta) ):

[ cos^2(theta) 1 - sin^2(theta) ]

Computing ( sin^2(theta) )

Substituting ( sin(theta) frac{8}{17} ) into the equation:

[ sin^2(theta) left(frac{8}{17}right)^2 frac{64}{289} ]

Substituting into the Pythagorean Identity

Now substitute ( sin^2(theta) ) into the identity:

[ cos^2(theta) 1 - frac{64}{289} ]

Simplifying the Right Side

Convert 1 to a fraction with a denominator of 289:

[ 1 frac{289}{289} ]

Thus, we have:

[ cos^2(theta) frac{289}{289} - frac{64}{289} frac{225}{289} ]

Finding ( cos(theta) )

Now take the square root to find ( cos(theta) ):

[ cos(theta) sqrt{frac{225}{289}} frac{sqrt{225}}{sqrt{289}} frac{15}{17} ]

Determining the Sign

Since ( theta arcsin(frac{8}{17}) ) is in the range (-frac{pi}{2}) to (frac{pi}{2}), where cosine is positive, we conclude:

[ cos(theta) frac{15}{17} ]

Conclusion

The exact value of ( cos(arcsin(frac{8}{17})) ) is:

[ boxed{frac{15}{17}} ]

Alternative Approaches

It's worth noting that this problem can also be approached by recognizing the Pythagorean triple 8–15–17. In this case:

- The given ( sin(theta) frac{8}{17} ) corresponds to the short side of the triangle.- The hypotenuse is 17, so the adjacent side must be 15 (since ( 17^2 - 8^2 289 - 64 225 15^2 )).

Therefore, ( cos(theta) frac{15}{17} ). This method is quicker but relies on recognizing the triple.