Video: Evaluating a Trigonometric Function Using a Trigonometric Identity

Given that cos (πœƒβ‚) = 1/3, where 0 < πœƒβ‚ < πœ‹/2 and cos (πœƒβ‚‚) = 1/3, where 0 < πœƒβ‚‚ < πœ‹/2, evaluate tan (πœƒβ‚ + πœƒβ‚‚) without using a calculator. Hint: Take tan (πœƒβ‚ + πœƒβ‚‚) = (tan (πœƒβ‚) + tan (πœƒβ‚‚))/(1 βˆ’ (tan (πœƒβ‚) tan (πœƒβ‚‚))).

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Video Transcript

Given that cos of πœƒ one is equal to one-third, where πœƒ one is between zero and πœ‹ over two and cos πœƒ two is equal to one-third, where πœƒ two is between zero and πœ‹ over two, evaluate tan of πœƒ one plus πœƒ two without using a calculator. Hint: Take tan of πœƒ one plus πœƒ two is equal to the tan of πœƒ one plus the tan of πœƒ two all over one minus the tan of πœƒ one times the tan of πœƒ two.

We’ve been given the cos of πœƒ one and the cos of πœƒ two, which are both one-third. In addition to that, both of the ranges for πœƒ one and πœƒ two are between zero and πœ‹ over two. In the context of a coordinate grid, if our angle falls between zero and two πœ‹, it will fall in quadrant one. And this gives us a bit more information about this angle. To remember this, we use the CAST diagram. The 𝐴 in quadrant one tells us that for angles that fall in quadrant one, all three of the trig functions will be positive. The sine, cosine, and tangent values of a πœƒ falling between zero and πœ‹ over two will be positive.

Since our goal is to find the tan of πœƒ one plus πœƒ two and we’ve been given a trig identity to help us do that, we need two pieces of information. We need to calculate the tan of πœƒ one and the tan of πœƒ two. We’ll need to know the sine, cosine, and tangent relationships. Since we’re given a cosine relationship, that is the adjacent angle over the hypotenuse and both πœƒ one and πœƒ two have the same cosine relationships. To calculate the tan of πœƒ one and πœƒ two, we need to know the opposite side length.

To help us find this opposite side length, we can sketch a right-angled triangle. Here’s a right-angled triangle with a cos of one-third. To calculate its opposite side length, we use the Pythagorean theorem, where we would say one squared plus 𝑏 squared equals three squared. One plus 𝑏 squared equals nine. If we subtract one from both sides, we see that 𝑏 squared equals eight. And to get 𝑏 by itself, we take the square root of both sides. We know the square root of eight is equal to the square root of four times two, which is equal to the square root of four times the square root of two. And so the most simplified form of 𝑏 is two times the square root of two.

We can add that to our right-angled triangle. And if we want to know the tangent relationship of this angle, it’s the opposite side length over the adjacent side length, two times the square root of two over one. Because both of these angles have a cos of one-third and fall in the same quadrant, the tan of πœƒ one and the tan of πœƒ two will be the same. Two times the square root of two over one. At this point, we’re ready to solve for the tan of πœƒ one plus πœƒ two.

We just need to plug in what we know. The tan of πœƒ one is two times the square root of two, and the tan of πœƒ two is two times the square root of two. We can add two times the square root of two plus two times the square root of two. When we combine them, we get four times the square root of two. In our denominator, we need to multiply two times the square root of two times two times the square root of two.

If we rearrange it, it could look like this. We multiply two times two, which gives us four. And the square root of two times the square root of two equals two. Four times two equals eight and one minus eight equals negative seven. So we can say the tan of πœƒ one plus πœƒ two is equal to negative four times the square root of two over seven.

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