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

this problem, we are asked to evaluate the expression the sine of the inverse tangent of 15 over 8. Now getting right into our steps here. In step 1, we're gonna take that inside function, and we're gonna use its interval to identify what quadrant our triangle will go in. Now here, my inside function is the inverse tangent and for the inverse tangent I know that my angles range from negative pi over 2 to positive pi over 2. So my triangle either has to go in quadrant 4 or quadrant 1 based on that interval. Now considering the sine of our argument here, we're taking the inverse tangent of a positive 15 over 8. This is a positive value. So based on my interval, I know that my tangent values can only be positive in quadrant 1. So that is where my triangle will go. Now that step 1 is done, moving on to step 2, we're actually going to draw our triangle and use our argument to label theta and our two sides. So drawing our triangle in that first quadrant here the way we have many times before, my triangle is situated laying against the x axis, and I can label my angle theta here measured from the x axis. So from here, I can use this argument 15 over 8. This tells me that the tangent of my angle theta is equal to that 15 over 8 to my argument there. And based on sohcahtoa, that tells me that my opposite side is going to be 15 and my adjacent side is 8. So completing step number 2, moving on to step 3, let's use our pythagorean theorem to find that missing third side. Now we're gonna set this up, a squared plus b squared equals c squared. We're trying to find our hypotenuse here, so we're solving for c squared. So plugging in my side lengths, 15 squared plus 8 squared gives me c squared, and 15 squared is 225. 8 squared is 64. That gives me c squared. Now adding these two values together gives me a value of 289. That's equal to c squared, the last step in finding this side length, taking the square root of both sides. Now the square root of 289 is 17, so that gives me my missing side length of 17. I can go ahead and label my hypotenuse on my triangle. Now we're done with step 3. Moving on to our last step here in getting our answer, we're gonna use our triangle to evaluate that outside function. Now here, my outside function is the sine. So in finding the sine of this angle theta based on SOHCAHTOA, I know that I need my opposite side, and I'm going to divide it by my hypotenuse. I have all the info I need. My opposite side is 15. My hypotenuse that we just found is 17 and this gives me my final answer. The sine of the inverse tangent of 15 over 8 is 15 over 17. Thanks for watching and I'll see you in the next one.

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Evaluate Composite Trig Functions

Trigonometry

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Evaluate Composite Trig Functions

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Multiple Choice

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

$\frac{8}{15}$

$\frac{15}{17}$

$\frac{8}{17}$

$\frac{15}{289}$

Verified step by step guidance

Recognize that $ \tan^{-1}\left(\frac{15}{8}\right) $ represents an angle $ \theta $ such that $ \tan(\theta) = \frac{15}{8} $.

Visualize or draw a right triangle where the opposite side to angle $ \theta $ is 15 and the adjacent side is 8. This helps in understanding the trigonometric relationships.

Use the Pythagorean theorem to find the hypotenuse of the triangle: $ c = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 $.

Now, find $ \sin(\theta) $ using the definition of sine in a right triangle: $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15}{17} $.

Thus, the expression $ \sin\left(\tan^{-1}\left(\frac{15}{8}\right)\right) $ evaluates to $ \frac{15}{17} $.