### Stony Brook MAT 123 Fall 2015 Trig values for special angles

 Start How to find the sine, cosine, tangent of special angles What do we mean by special angles? Well if you recall From geometry, An equilateral triangle has three 60 degree triangles If we were to drop a perpendicular from any of those 60 degree angles To the opposite side it cuts the vertex angle in half and these are both 30 degree angles 0:32 and what do we know about an equilateral triangle? Well all the sides are equal, so if we call these sides 2x Then this distance is x because this perpendicular is a perpendicular bisector and cuts this base in half So, with a little pythagorean theorem you can now find that this distance is x times the square root of 3. 1:00 So if we took it out of the equilateral triangle We find that we have a, whats called a 30, 60, 90 triangle. The side opposite the 30 degrees, that side is x. The side opposite the 60 degrees is x radical 3, thats the length and then the hypotenuse is 2x. 1:30 So now let's find the sine of 30 degrees Well the sine of 30 degrees is the opposite side divided by the hypotenuse Which is x divided by 2x. Use the miracle of algebra and cancel the x's and you get a half The cosine of 30 degrees, is the adjacent side divided by 2x and that equals the square root of 3 over 2 2:03 And the tangent of 30 degrees, is x over x times the square root of 3 or 1 over the square root of 3. The sine of 60 degrees, is the side opposite 60 degrees, x radical 3, over 2x. 2:30 Which simplifies to the square root of 3 over 2 The cosine of 60 degrees is the adjacent side, x over 2x Which simplifies to a half And the tangent of 60 degrees is opposite over adjacent, is x radical 3 over x Which simplifies to the square root of 3. 3:02 Okay, now let's do another triangle! Do you remember from a square? If you cut it on a diagonal, both of those angles are now 45 degrees. 3:32 If these sides are the same, they're both x. And this, with a little pythagorean theorem work, you can show it's x times the square root of 2. So now we can find the sine and the cosine. So let's take this triangle out.. 4:01 These are both 45 degrees So the sine of 45 degrees is x over x radical 2. Which simplifies to 1over the square root of 2. And if you rationalize the denominator, you'll get that that can converted to the square root of 2 over 2. Either of those is the same of course. The cosine of 45 degrees, is the adjacent over the hypotenuse. It's also 1 over the square root of 2. 4:39 And the tangent of 45 degrees is x over x, which is 1. So now let's finally come up with an easy way to memorize these. These triangles, sine, cosine, and tangent will be very useful throughout the course 5:00 because we will use them over and over again, since you don't need a calculator to figure them out So an easy way to memorize them is make a box That's 3 by 3. You write sine, cosine, and tangent down the side and 30 degrees, 45 degrees and 60 degrees across the top. Then we are going to need some fraction bars 5:33 Sine and cosine, everything is over 2. and then you remember for sine.. it goes 1, 2, 3. So, this is 1, square root of 2, square root of 3. Then cosine just goes in the other direction. It goes 3, 2, 1. and then tangent you find by taking the sines numerator and divided it by the cosine numerator. 6:02 because you really are taking the sine dividing it by the cosine and when you do, the denominators cancel The sine over the cosine here comes out 1. and here you get the square root of 3. And those are the special triangles!