Question Video: Using Inverse Trigonometric Functions to Evaluate Trigonometric Expressions

Find the exact value of tan⁻¹ (tan(βˆ’5πœ‹/6)) in radians.

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Video Transcript

Find the exact value of tan inverse of tan of negative five πœ‹ over six in radians.

If you’re lucky, putting this expression into your calculator, assuming it’s in radian mode, will give you an answer of πœ‹ over six; you’ll get the exact value for free. If your calculator doesn’t give you the exact value, you’ll get a decimal approximation of it, which is around 0.52. Either way, you should be able to see that the answer your calculator gives you is not negative five πœ‹ over six.

And this might be a surprise because you’d think that if you take a number, apply a function to it, and then apply the inverse function of that function to the result, you should get the number you started with. But of course, in this case, we don’t; we get πœ‹ over six instead. The rest of this video will be about explaining why this is the case; why we get πœ‹ by six and not negative five πœ‹ over six?

Have a look at the graph of 𝑦 equals tan π‘₯. We can use it to read off the value of tan negative five πœ‹ over six. This value appears to be just a bit greater than 0.5. There is an exact value, which your calculator may give you of root three over three, which is approximately 0.577. And so our question about the inverse tan of tan of negative five πœ‹ over six becomes a question about the inverse tan of root three over three, which is not 0.577 and so on.

Inverse tan is a function which tries to undo the tangent function. And so when asking for the tan inverse of 0.577 dot dot dot, what we’re asking for is a value of πœƒ, for which tan πœƒ is 0.577 dot dot dot. If we look at our graph, we can see that unfortunately there are lots of values of π‘₯, for which tan has this value. In fact, as the tan function is periodic, there are infinitely many values of π‘₯, for which tan π‘₯ is 0.577 dot dot dot.

Inverse tan is a function. And so it can only return one of those numerical values. How does it decide which value to return? The trick is that inverse tan always returns a value between negative πœ‹ by two and πœ‹ by two. There is a unique value of π‘₯ in this region, for which tan of π‘₯ is 0.577 dot dot dot. This value is πœ‹ by six. And we can see in this region that there is a unique value of π‘₯, which corresponds to any given value of tan π‘₯. There is, for example, a unique value of π‘₯, for which tan π‘₯ is negative two. So the tan inverse of 0.377 dot dot dot is πœ‹ by six.

The other values of π‘₯, for which tan π‘₯ is root three over three or 0.577 dot dot dot, can be found easily from this inverse tangent value. π‘₯ could be πœ‹ by six as discussed. But as tan is periodic with periods πœ‹, it could also be πœ‹ by six plus πœ‹ or seven over six times πœ‹. Similarly, if we subtract πœ‹ instead from our inverse tangent value, we get negative five πœ‹ over six, which was a value we were expecting. And we can keep going for as long as we want; the tangent of negative 11πœ‹ over six is also root three over three or 0.577 dot dot dot. All of these values of π‘₯ are separated by πœ‹, so they are all multiples of πœ‹ around for my inverse tangent value, one over six πœ‹.

And so if you wanted to find the exact value of tan inverse of tan of π‘₯, for some value of π‘₯, well, if π‘₯ is in the region between negative πœ‹ by two and πœ‹ by two, then you’re in luck. The inverse tan of tan of π‘₯ is then just π‘₯; otherwise, you will have to either add or subtract a multiple of πœ‹ to get a value in this region. And then that value will be your answer.

So, for example, let’s go back to our original question: tan inverse of tan of negative five πœ‹ over six. We can see that negative five πœ‹ over six is not between negative πœ‹ by two and πœ‹ by two, but we can add a multiple of πœ‹, in this case, just πœ‹. And then, the value will be in this range. And so the inverse tan of the tan of negative five πœ‹ over six is this value negative five πœ‹ by six plus πœ‹, which is πœ‹ by six.

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