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Published byJaylin Marsland Modified over 2 years ago

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Right Triangles Consider the following equilateral triangle, with each side having a value of 2. Drop a perpendicular segment from the top vertex to the base side. The base side has been bisected into two segments of length 1.

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**Since the original triangle was equilateral, the base angles are 60° each.**

Since the angle at the top of the triangle has been bisected, the original 60° angle has been split into new angles that are 30° each.

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**Consider the new triangle formed on the left side.**

This is a right triangle. We have just proven that in a triangle the short leg (always opposite the 30° angle) is always half the length of the hypotenuse.

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**Name the long leg b and determine its value.**

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This right triangle can give us the trigonometric function values of 30° and 60°. We first do the 30° angle.

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We now do the 60° angle.

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**This leads us to some important values on the unit circle.**

Recall that on the unit circle we have …

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**Consider the point (a,b) on the 30° ray of a unit circle.**

Since (a,b) = (cos 30°, sin 30°) , we have

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**In radian form it would be …**

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**Moving around the unit circle with reference angles of π/6 we have …**

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Example 1: Find cos 5π/6 Since cos x is equal to the first coordinate of the point we have …

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Example 2: Find sin 7π/6 Since sin x is equal to the second coordinate of the point we have …

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Example 3: Find tan (-7π/6) Since tan x is equal to b/a we have …

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**In similar fashion, moving around the unit circle with reference angles of π/3 (60°) we have …**

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Example 1: Find cos 4π/3 Since cos x is equal to the first coordinate of the point we have …

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Example 2: Find sin -2π/3 Since sin x is equal to the second coordinate of the point we have …

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Example 3: Find tan (2π/3) Since tan x is equal to b/a we have …

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