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