Trigonometrical Pythagorean Identity - SS3 Mathematics Lesson Note

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Given the triangle below:

By Pythagoras theorem, \(r^{2} = x^{2} + y^{2}\)

By trigonometrical ratios,

\(\cos\theta = \frac{x}{r}\); \(x = r\cos\theta\)

\(\sin\theta = \frac{y}{r}\); \(y = r\sin\theta\)

\[x^{2} = r^{2}\cos^{2}\theta\]

\[y^{2} = r^{2}\sin^{2}\theta\]

Combining Pythagorean Theorem and trigonometric ratios,

\[r^{2} = x^{2} + y^{2}\]

\[r^{2} = r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta\]

\[r^{2} = r^{2}(\cos^{2}\theta + \sin^{2}\theta)\]

\[\cos^{2}\theta + \sin^{2}\theta = \frac{r^{2}}{r^{2}}\]

\[\cos^{2}\theta + \sin^{2}\theta = 1\]

\(\mathbf{TRIGONOMETRICAL\ PYTHAGOREAN\ IDENTITY\ 1:\ }\mathbf{\cos}^{\mathbf{2}}\mathbf{\theta}\mathbf{+}\mathbf{\sin}^{\mathbf{2}}\mathbf{\theta}\mathbf{= 1}\), which holds true for all values of \(\mathbf{\theta}\).

Given \(\cos^{2}\theta + \sin^{2}\theta = 1\), dividing through by \(\cos^{2}\theta\)

\[\frac{\cos^{2}\theta}{\cos^{2}\theta} + \frac{\sin^{2}\theta}{\cos^{2}\theta} = \frac{1}{\cos^{2}\theta}\]

\[1 + \tan^{2}{\theta = \sec^{2}\theta}\]

\(\mathbf{TRIGONOMETRICAL\ PYTHAGOREAN\ IDENTITY\ 2:\ 1 +}\mathbf{\tan}^{\mathbf{2}}{\mathbf{\theta =}\mathbf{\sec}^{\mathbf{2}}\mathbf{\theta}}\), which holds true for all values of \(\mathbf{\theta}\).

Given \(\cos^{2}\theta + \sin^{2}\theta = 1\), dividing through by \(\sin^{2}\theta\)

\[\frac{\cos^{2}\theta}{\sin^{2}\theta} + \frac{\sin^{2}\theta}{\sin^{2}\theta} = \frac{1}{\sin^{2}\theta}\]

\[\cot^{2}{\theta + 1 = {cosec}^{2}\theta}\]

\(\mathbf{TRIGONOMETRICAL\ PYTHAGOREAN\ IDENTITY\ 3:\ }\mathbf{\cot}^{\mathbf{2}}{\mathbf{\theta + 1 =}\mathbf{cosec}^{\mathbf{2}}\mathbf{\theta}}\), which holds true for all values of \(\mathbf{\theta}\).

Example 1 If \(x = 2r\cos\theta\) and \(y = 2r\sin\theta\), find \(\sqrt{x^{2} + y^{2}}\)

Solution

\[\sqrt{x^{2} + y^{2}} = \ \sqrt{{(2r\cos\theta)}^{2} + {(2r\sin\theta)}^{2}}\]

\[= \ \sqrt{4r^{2}\cos^{2}\theta + 4r^{2}\sin^{2}\theta}\]

\[= \ \sqrt{4r^{2}{\ (cos}^{2}\theta + \sin^{2}\theta)}\]

\[= \sqrt{4r^{2}}.\sqrt{{\ (cos}^{2}\theta + \sin^{2}\theta)} = 2r.\sqrt{1} = 2r\]

Example 2 Find the value of \(\theta\) between \(0{^\circ}\) and \(360{^\circ}\) which satisfies the trigonometric equation \(\sin^{2}\theta + \sin\theta = 0\)

Solution

\[\sin^{2}\theta + \sin\theta = 0\]

Let \(P = \sin\theta\)

\[P^{2} + P = 0\]

Factorizing: \(P(P + 1) = 0\)

Thus, \(P = 0\) or \(P + 1 = 0\)

\(P = 0\) or \(P = - 1\)

\(\sin\theta = 0\) or \(\sin\theta = - 1\)

\[\theta = \sin^{- 1}0 = 0{^\circ}\ or\ 180{^\circ}\ or\ 360{^\circ}\]

\[\theta = \sin^{- 1}{( - 1)} = 270{^\circ}\]

Thus, \(\theta = 0{^\circ},\ 180{^\circ},270{^\circ}\ or\ 360{^\circ}\)