06Ecomplementary

= Complementary Functions =
 * Recall that two angles are ** complementary ** if they add to 90º


 * Recall that in a right-angled triangle, the two non-right angles are complementary.


 * Consider the right angled triangle within the unit circle
 * the hypotenuse is 1 (because the radius is 1)
 * the side opposite to q is defined as sin(q )
 * the side adjacent to q is defined as cos(q )


 * The angle at A must be the complement of q
 * So
 * Angle A = 90º – q
 * In radians:
 * A = p /2 – q


 * Applying SOHCAHTOA to the angle A, we get:

math . \qquad \qquad \sin \big( 90^\circ - \theta \big) = \dfrac{OPP}{HYP} \\.\\ . \qquad \qquad \sin \big( 90^\circ - \theta \big) = \dfrac{ \cos \big( \theta \big) }{1} \\.\\ . \qquad \qquad \sin \big( 90^\circ - \theta \big) = \cos \big( \theta \big) math


 * And (similarly)

math . \qquad \qquad \cos \big( 90^\circ - \theta \big) = \sin \big( \theta \big) math




 * Or, in radians:


 * What this means is that
 * sin(10º) = cos(80º)
 * sin(25º) = cos(65º)
 * sin(60º) = cos(30º)
 * etc


 * Because of this connection, sine and cosine are called ** Complementary Functions **


 * We will use these rules in the first quadrant, but it is possible to extend them into the other quadrants as well


 * Example 1 **

... ... Given that cos(a) = 0.7 .. and a is in Q1 ... ... Find sin(90º – a)


 * Solution:**

... ... sin(90º – a) = 0.7


 * Example 2 **

math . \qquad \text{Given that } \sin \Big( \dfrac{\pi}{10} \Big) = \dfrac{\sqrt{5}-1}{4} \\.\\ . \qquad \text{Find} \\.\\ . \qquad \textbf{(a)} \quad \cos \Big( \dfrac{2\pi}{5} \Big) \\.\\ . \qquad \textbf{(b)} \quad \sin \Big( \dfrac{2\pi}{5} \Big) math


 * Solution:**

math . \qquad \dfrac{2\pi}{5} = \dfrac{4\pi}{10} \\.\\ . \qquad \dfrac{2\pi}{5} = \dfrac{5\pi}{10}-\dfrac{\pi}{10} \\.\\. \\ . \qquad \textbf{(a)} \quad \cos \Big( \dfrac{2\pi}{5} \Big) \\.\\ . \qquad \qquad \qquad = \cos \Big( \dfrac{\pi}{2} - \dfrac{\pi}{10} \Big) \\.\\ . \qquad \qquad \qquad = \sin \Big( \dfrac{\pi}{10} \Big) \\.\\ . \qquad \qquad \qquad = \dfrac{ \sqrt{5} - 1}{4} math

math . \qquad \textbf{(b)} \quad \sin \Big( \dfrac{2\pi}{5} \Big) math
 * Draw a right angled triangle using this information
 * Use Pythagoras' Theorem to find the missing side

math . \qquad \text{OPP}^2 = 4^2 - \big( \sqrt{5} - 1 \big)^2 \\.\\ . \qquad \text{OPP}^2 = 16 - \big( 5 - 2\sqrt{5} + 1 \big) \\.\\ . \qquad \text{OPP}^2 = 10 + 2\sqrt{5} \\.\\ . \qquad \text{OPP} = \sqrt{ 10 + 2 \sqrt{5} } math


 * Apply SOHCAHTOA

math . \qquad \sin \Big( \dfrac{2\pi}{5} \Big) = \dfrac{ \sqrt{ 10 + 2\sqrt{5}}}{4} math

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