06Etrigidentities

= Trigonometric Identities =


 * An ** identity ** is like an equation but it is true for all values for which it is defined


 * For example,
 * x 2 + 3x = x ( x + 3 ) ... is true for all values of x


 * The Pythagorean Identity **


 * Recall the diagram where we defined sine and cosine within the unit circle
 * Notice that we have a right-angled triangle within that diagram
 * Apply Pythagoras' Theorem and we get:

math . \qquad a^2 + b^2 = c^2 \\.\\ . \qquad \sin^2 \big( \theta \big) + \cos^2 \big( \theta \big) = 1 math


 * Note, we write sin 2 (q ) instead of sin(q ) 2
 * to avoid confusing it with sin ( q 2 )
 * and to save us writing ( sin(q ) ) 2


 * This is called the ** Pythagorean Identity ** because
 * it is true for all values of q
 * it came from Pythagoras' Theorem (doh!)


 * Example 1 **

... ... If sin(a ) = 0.9 and 90º < a < 180º ... ... Find cos(a ) ... (2 decimal places)


 * Solution:**

math . \qquad \sin^2 \big( \alpha \big) + \cos^2 \big( \alpha \big) = 1 \\.\\ . \qquad \big( 0.9 \big)^2 + \cos^2 \big( \alpha \big) = 1 \\.\\ . \qquad 0.81 + \cos^2 \big( \alpha \big) = 1 \\. math math . \qquad \cos^2 \big( \alpha \big) = 0.19 \\.\\ . \qquad \cos \big( \alpha \big) = \pm \sqrt{0.19} \\.\\ . \qquad \cos \big( \alpha \big) = \pm 0.44 \\.\\ math
 * a is in 2nd Quadrant
 * cos is negative in Q2, so

math . \qquad \cos \big( \alpha \big) = -0.44 math


 * Example 2 **

math . \qquad \text{If } \cos\big(\beta\big) = -\dfrac{\sqrt{5}}{4} \quad \text{and} \quad \beta \in \Big[ \pi, \dfrac{3\pi}{2} \Big] \\.\\ . \qquad \text{Find } \sin \big( \beta \big) math


 * Solution;**

math . \qquad \sin^2 \big( \beta \big) + \cos^2 \big( \beta \big) = 1 \\.\\ . \qquad \sin^2 \big( \beta \big) + \Big( -\dfrac{\sqrt{5}}{4} \Big)^2 = 1 \\.\\ . \qquad \sin^2 \big( \beta \big) + \dfrac{5}{16} = 1 \\. math math . \qquad \sin^2 \big( \beta \big) = \dfrac{11}{16} \\.\\ . \qquad \sin \big( \beta \big) = \pm \dfrac{\sqrt{11}}{4} \\.\\ math
 * b is in 3rd Quadrant
 * sin is negative in Q3, so

math . \qquad \sin \big( \beta \big) = -\dfrac{\sqrt{11}}{4} math


 * The Tangent Identity **


 * Recall that from SOHCAHTOA we get that

math . \qquad \qquad \tan \big( \theta \big) = \dfrac{OPP}{ADJ} math


 * Look at the same diagram for a unit circle
 * Apply the tan rule to the right angled triangle and we get:

math . \qquad \tan \big( \theta \big) = \dfrac{ \sin \big( \theta \big) }{ \cos \big( \theta \big) } math


 * This is called the ** Tangent Identity ** because
 * it is true for all values of q where tan(q ) is defined

math . \qquad \tan \big( \theta \big) \; \text{ is not defined when } \; \cos \big( \theta \big) = 0 \\.\\ . \qquad \cos \big( \theta \big) = 0 \; \text{ when } \; \theta = \Big\{ ... \dfrac{\pi}{2}, \; \dfrac{3\pi}{2}, \; \dfrac{5\pi}{2}, \; ... \Big\} \\.\\ math

math . \qquad \text{hence} \\.\\ . \qquad \tan \big( \theta \big) \; \text{ is not defined when } \; \theta = \dfrac{ \big( 2n+1 \big) \pi}{2} \quad n \in Z \; \big\{ integers \big\} math

Example 3

... ... If sin(a ) = 0.9 and cos(a ) = –0.44 and 90º < a < 180º ... ... Find tan(a ) ... (2 decimal places)


 * Solution:**

math . \qquad \tan\big( \alpha \big) = \dfrac{ \sin \big( \alpha \big) }{ \cos \big( \alpha \big) } \\.\\ . \qquad \tan\big( \alpha \big) = \dfrac{ 0.9 }{ -0.44 } \\.\\ . \qquad \tan\big( \alpha \big) = -2.05 math


 * Note tan is negative in Q2, no adjustment needed to answer


 * Example 4 **

math . \qquad \text{If } \; \sin \big( \beta \big) = -\dfrac{\sqrt{11}}{4} \\.\\ . \qquad \text{and } \cos\big(\beta\big) = -\dfrac{\sqrt{5}}{4} \\.\\ . \qquad \text{and} \quad \beta \in \Big[ \pi, \dfrac{3\pi}{2} \Big] \\.\\ . \qquad \text{Find } \tan \big( \beta \big) math


 * Solution:**

math . \qquad \tan\big( \beta \big) = \dfrac{ \sin \big( \beta \big) }{ \cos \big( \beta \big) } \\.\\ . \qquad \tan\big( \beta \big) = -\dfrac{\sqrt{11}}{4} \div -\dfrac{\sqrt{5}}{4} \\.\\ . \qquad \tan\big( \beta \big) = -\dfrac{\sqrt{11}}{4} \times -\dfrac{4}{\sqrt{5}} \\.\\ . \qquad \tan\big( \beta \big) = \dfrac{\sqrt{11}}{\sqrt{5}} \qquad \Bigg\{ \times \dfrac{ \sqrt{5}}{\sqrt{5}} \Bigg\}\\.\\ . \qquad \tan\big( \beta \big) = \dfrac{\sqrt{55}}{5} math


 * Note tan is positive in Q3, no adjustment needed to answer


 * Alternate Method **


 * Use the given information to construct a right-angled triangle (ignoring signs)
 * Use Pythagoras to fill in the missing side
 * Use SOHCAHTOA to find the value of the trig functions requested
 * Apply plus or minus depending on the quadrant


 * Example 5 **

math . \qquad \text{Given that }\; \cos \big( \theta \big) = -\dfrac{ \sqrt{7}}{3} \quad \theta \in \Big( \dfrac{\pi}{2}, \; \pi \Big) \\.\\ . \qquad \text{Find} \\.\\ . \qquad \textbf{(a)} \quad \sin \big( \theta \big) \\.\\ . \qquad \textbf{(b)} \quad \tan \big( \theta \big) math


 * Solution:**
 * Use the given information to construct a right-angled triangle

math . \qquad \cos \big( \theta \big) = -\dfrac{ \sqrt{7}}{3} \quad \text{so} \\.\\ . \qquad \qquad \text{ADJ } = \sqrt{7} \\.\\ . \qquad \qquad \text{HYP } = 3 math


 * Use Pythagoras to find the missing side

math . \qquad \text{OPP}^2 = 3^2 - \big( \sqrt{7} \big)^2 \\.\\ . \qquad \text{OPP}^2 = 9 - 7 = 2 \\.\\ . \qquad \text{OPP } = \sqrt{2} math


 * Use SOHCAHTOA to find the value of the trig functions requested
 * In Q2, so
 * sin(<span style="font-family: Symbol,sans-serif; font-size: 120%;">q ) is positive
 * tan(<span style="font-family: Symbol,sans-serif; font-size: 120%;">q ) is negative

math . \qquad \textbf{(a)} \quad \sin \big( \theta \big) = +\dfrac{ \sqrt{2}}{3} \\.\\ . \qquad \textbf{(b)} \quad \tan \big( \theta \big) = -\dfrac{ \sqrt{2}}{ \sqrt{7}} = -\dfrac{ \sqrt{14}}{7} math

math . \qquad \textbf{Note: } \text{ Rationalise denominator of answer } \textbf{(b)} \\ . \qquad \qquad \quad \text{by multiplying top and bottom by } \sqrt{7} math .