2CPythagIdents

Pythagorean Identities

We know that for a right angled triangle, with hypotenuse equal to 1, we get:

Using Pythagoras, this gives us:

math . \qquad \sin^2 \theta + \cos^2 \theta = 1 math

If we divide both sides by cos 2 q, we get:

math . \qquad \dfrac {\sin^2 \theta}{\cos^2 \theta} + \dfrac {\cos^2 \theta}{\cos^2 \theta} = \dfrac {1}{\cos^2 \theta} math . math . \qquad \tan^2 \theta + 1 = \sec^2 \theta \qquad \theta \in R \backslash \Big\{ \dfrac {\pi}{2} + n\pi, \;\; n \in Z \Big\} math

{values excluded because cos = 0 makes both tan and sec undefined}

OR
If we divide both sides by sin 2 q, we get:

math . \qquad \dfrac {\sin^2 \theta}{\sin^2 \theta} + \dfrac {\cos^2 \theta}{\sin^2 \theta} = \dfrac {1}{\sin^2 \theta} math . math . \qquad 1 + \cot^2 \theta = \text{cosec}^2 \theta \qquad \theta \in R \backslash \{ n\pi, \;\; n \in Z \} math

{values excluded because sin = 0 makes both cot and cosec undefined}

Example 1 math \text {Given } \tan(x) = 2, \quad x \in \left[0, \; \dfrac {\pi}{2} \right] math find sec(x), cos(x), sin(x), cosec(x)

math \\ . \qquad \text {Use } \sec^2 \theta = \tan^2 \theta + 1 \\. \\ . \qquad \sec^2(x) = 4 + 1 = 5 \\. \\ . \qquad \sec(x) = \pm \sqrt{5} math
 * a) sec(x) **

but domain in first quadrant where cos is positive, so sec is positive

math . \qquad \sec(x) = \sqrt{5} math

math \\ . \qquad \text {Use } \cos \theta = \dfrac {1}{\sec \theta} \\. \\ . \qquad \cos (x) = \dfrac {1}{\sqrt {5}} = \dfrac {\sqrt{5}}{5} math
 * b) cos(x) **

math \\ . \qquad \text{Use } \sin^2 \theta + \cos^2 \theta = 1 \\. \\ . \qquad \sin^2 (x) + \left( \dfrac {1}{\sqrt{5}} \right)^2 = 1 \\. \\ . \qquad \sin^2 (x) + \dfrac {1}{5} = 1 \\. \\ . \qquad \sin^2 (x) = \dfrac {4}{5} \\. \\ . \qquad \sin (x) = \pm \dfrac {2}{\sqrt{5}} = \pm \dfrac {2\sqrt{5}}{5} math
 * c) sin(x) **

But in first quadrant, so sin is positive

math . \qquad \sin (x) = \dfrac {2\sqrt{5}}{5} math

math \\ . \qquad \text {Use } \text{cosec } \theta = \dfrac {1}{\sin \theta} \\. \\ . \qquad \text {cosec} (x) = \dfrac {\sqrt {5}}{2} math
 * d) cosec(x) **

Example 2

math \text{Simplify the expression : } \; \dfrac {\cos x - \cos^3 x}{\cot x} math

Change cot to 1/tan and then change tan to sin/cos

math \\ . \qquad = (\cos x - \cos^3 x)\tan x \\. \\ . \qquad = \cos x (1 - \cos^2 x)\tan x \\. \\ . \qquad = \cos x (\sin^2 x) \dfrac {\sin x}{\cos x} \\. \\ . \qquad = \sin^3 x math

For another site that explains this idea, go here: MathsIsFun

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