1.5.3 Trigonometric Identities - GCEALevelMaths9709

In this topic we will learn how to:

use the identities $\frac{\sin{\theta}}{\cos{\theta}} \equiv \tan{\theta}$ and $\sin^{2}{\theta} + \cos^{2}{\theta} \equiv 1$

$\textbf{\large\textcolor{gray}{Pythagorean Identities}}$ The two trig identities (also known as pythagorean identities) displayed below will be used in proving other trig identities,
$\tan{\theta} \equiv \frac{\sin{\theta}}{\cos{\theta}}$ $\sin^{2}{\theta} + \cos^{2}{\theta} \equiv 1$ In some cases, the second trig identity, may be useful in the forms below,
$\sin^{2}{\theta} \equiv 1 - \cos^{2}{\theta}$ $\cos^{2}{\theta} \equiv 1 - \sin^{2}{\theta}$ General tips for solving trig identities,

Work from the more complex side to the less complex side
Reduce anything in terms of $\tan{x}$ to be in terms of $\sin{x}$ and $\cos{x}$
Combine any separate fractions into one single fraction
Expose yourself to a lot trig identity questions

Let’s look at some past paper questions.

1. Show that,

$\frac{\sin{\theta} + 2\cos{\theta}}{\cos{\theta} - 2\sin{\theta}} - \frac{\sin{\theta} - 2\cos{\theta}}{\cos{\theta} + 2\sin{\theta}} \equiv \frac{4}{5\cos^{2}{\theta} - 4}$

(9709/12/F/M/22 number 7)

We will work from the left-hand side to the right-hand side,
$\frac{\sin{\theta} + 2\cos{\theta}}{\cos{\theta} - 2\sin{\theta}} - \frac{\sin{\theta} - 2\cos{\theta}}{\cos{\theta} + 2\sin{\theta}}$ Compose the two fractions into one fraction,

$\frac{(\sin{\theta} + 2\cos{\theta})(\cos{\theta} + 2\sin{\theta}) - (\sin{\theta} - 2\cos{\theta})(\cos{\theta} - 2\sin{\theta})}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$

Expand the numerator,

$\frac{\sin{\theta} \cos{\theta} + 2\sin^{2}{\theta} + 2\cos^{2}{\theta} + 4\sin{\theta} \cos{\theta} - (-2\sin^{2}{\theta} + \sin{\theta} \cos{\theta} - 2\cos^{2}{\theta} + 4\sin{\theta} \cos{\theta})}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$

Simplify the numerator,

$\frac{5\sin{\theta} \cos{\theta} + 2\sin^{2}{\theta} + 2\cos^{2}{\theta} - (-2\sin^{2}{\theta} - 2\cos^{2}{\theta} + 5\sin{\theta} \cos{\theta})}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$

$\frac{5\sin{\theta} \cos{\theta} + 2(\textcolor{#2192ff}{\sin^{2}{\theta} + \cos^{2}{\theta}}) - (-2(\textcolor{#2192ff}{\sin^{2}{\theta} + \cos^{2}{\theta}}) + 5\sin{\theta} \cos{\theta})}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$

Use the identity $\textcolor{#2192ff}{\sin^{2}{\theta} + \cos^{2}{\theta} \equiv 1}$ ,
$\frac{5\sin{\theta} \cos{\theta} + 2(\textcolor{#2192ff}{1}) - (-2(\textcolor{#2192ff}{1}) + 5\sin{\theta} \cos{\theta})}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$ $\frac{5\sin{\theta} \cos{\theta} + 2 - (-2 + 5\sin{\theta} \cos{\theta})}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$ Finish expanding the numerator,
$\frac{5\sin{\theta} \cos{\theta} + 2 + 2 - 5\sin{\theta} \cos{\theta}}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$ Finish simplifying the numerator,
$\frac{4}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}$ Expand the denominator,
$\frac{4}{\cos^{2}{\theta} + 2\sin{\theta} \cos{\theta} - 2\sin{\theta} \cos{\theta} - 4\sin^{2}{\theta}}$ Simplify the denominator,
$\frac{4}{\cos^{2}{\theta} - 4\textcolor{#2192ff}{\sin^{2}{\theta}}}$ Use the identity $\textcolor{#2192ff}{\sin^{2}{x} \equiv 1 - \cos^{2}{x}}$ in the denominator,
$\frac{4}{\cos^{2}{\theta} - 4(\textcolor{#2192ff}{1 - \cos^{2}{\theta}})}$ Simplify the denominator,
$\frac{4}{\cos^{2}{\theta} - 4 + 4\cos^{2}{\theta}}$ Group like terms in the denominator, $\frac{4}{\cos^{2}{\theta} + 4\cos^{2}{\theta} - 4}$ $\frac{4}{5\cos^{2}{\theta} - 4}$ Therefore,
$\frac{\sin{\theta} + 2\cos{\theta}}{\cos{\theta} - 2\sin{\theta}} - \frac{\sin{\theta} - 2\cos{\theta}}{\cos{\theta} + 2\sin{\theta}} \equiv \frac{4}{5\cos^{2}{\theta} - 4}$ 2. Prove the identity,

$\frac{1 + \sin{x}}{1 - \sin{x}} - \frac{1 - \sin{x}}{1 + \sin{x}} \equiv \frac{4\tan{x}}{\cos{x}}$

(9709/12/M/J/21 number 10)

We will work from the left-hand side to the right-hand side,
$\frac{1 + \sin{x}}{1 - \sin{x}} - \frac{1 - \sin{x}}{1 + \sin{x}}$ Compose the two fractions into one fraction,
$\frac{(1 + \sin{x})^{2} - (1 - \sin{x})^{2}}{(1 - \sin{x})(1 + \sin{x})}$ Expand the numerator,
$\frac{1 + 2\sin{x} + \sin^{2}{x} - (1 - 2\sin{x} + \sin^{2}{x})}{(1 - \sin{x})(1 + \sin{x})}$ $\frac{1 + 2\sin{x} + \sin^{2}{x} - 1 + 2\sin{x} - \sin^{2}{x}}{(1 - \sin{x})(1 + \sin{x})}$ Group like terms and simplify the numerator,
$\frac{1 - 1 + 2\sin{x} + 2\sin{x} + \sin^{2}{x} - \sin^{2}{x}}{(1 - \sin{x})(1 + \sin{x})}$ $\frac{4\sin{x}}{(1 - \sin{x})(1 + \sin{x})}$ Expand the denominator,
$\frac{4\sin{x}}{\textcolor{#2192ff}{1 - \sin^{2}{x}}}$ Use the identity $\textcolor{#2192ff}{\cos^{2}{x} \equiv 1 - \sin^{2}{x}}$ in the denominator,
$\frac{4\sin{x}}{\textcolor{#2192ff}{\cos^{2}{x}}}$ Simplify,
$\frac{4\sin{x}}{\cos{x} \times \cos{x}}$ $\frac{4\sin{x}}{\cos{x}} \times \frac{1}{\cos{x}}$ $4\tan{x} \times \frac{1}{\cos{x}}$ $\frac{4\tan{x}}{\cos{x}}$ Therefore,
$\frac{1 + \sin{x}}{1 - \sin{x}} - \frac{1 - \sin{x}}{1 + \sin{x}} \equiv \frac{4\tan{x}}{\cos{x}}$ 3. Prove the identity,

$\frac{1 - 2\sin^{2}{\theta}}{1 - \sin^{2}{\theta}} = 1 - \tan^{2}{\theta}$

(9709/11/M/J/21 number 7)

We will work from the right-hand side to the left-hand side,
$1 - \textcolor{#2192ff}{\tan^{2}{\theta}}$ Note: In this case, you could just as easily work from left-hand side to right-hand side, however, I chose the opposite because it is easier to simplify $\tan^{2}{\theta}$ to be in terms of $\sin{\theta}$ and $\cos{\theta}$ .

Use the identity $\textcolor{#2192ff}{\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}}$ ,
$1 - \textcolor{#2192ff}{\frac{\sin^{2}{\theta}}{\cos^{2}{\theta}}}$ Compose the two terms into one fraction,
$\frac{\textcolor{#0f0}{\cos^{2}{\theta}} - \sin^{2}{\theta}}{\cos^{2}{\theta}}$ Use the identity $\textcolor{#0f0}{\cos^{2} \equiv 1 - \sin^{2}{\theta}}$ in the numerator,
$\frac{\textcolor{#0f0}{1 - \sin^{2}{\theta}} - \sin^{2}{\theta}}{\cos^{2}{\theta}}$ Simplify the numerator,
$\frac{1 - 2\sin^{2}{\theta}}{\textcolor{red}{\cos^{2}{\theta}}}$ Use the identity $\textcolor{red}{\cos^{2} \equiv 1 - \sin^{2}{\theta}}$ in the denominator,
$\frac{1 - 2\sin^{2}{\theta}}{\textcolor{red}{1 - \sin^{2}{\theta}}}$ Therefore,
$\frac{1 - 2\sin^{2}{\theta}}{1 - \sin^{2}{\theta}} = 1 - \tan^{2}{\theta}$ 4. Prove the identity,

$\frac{\sin^{3}{\theta}}{\sin{\theta} - 1} - \frac{\sin^{2}{\theta}}{1 + \sin{\theta}} \equiv -\tan^{2}{\theta}(1 + \sin^{2}{\theta})$

(9709/11/M/J/22 number 4)

We will work from the left-hand side to the right-hand side,
$\frac{\sin^{3}{\theta}}{\sin{\theta} - 1} - \frac{\sin^{2}{\theta}}{1 + \sin{\theta}}$ Compose the two fractions into one fraction,
$\frac{\sin^{3}{\theta}(1 + \sin{\theta}) - \sin^{2}{\theta}(\sin{\theta} - 1)}{(\sin{\theta} - 1)(1 + \sin{\theta})}$ Expand the numerator and simplify,
$\frac{\sin^{3}{\theta} + \sin^{4}{\theta} - \sin^{3}{\theta} + \sin^{2}{\theta}}{(\sin{\theta} - 1)(1 + \sin{\theta})}$ $\frac{\sin^{4}{\theta} + \sin^{2}{\theta}}{(\sin{\theta} - 1)(1 + \sin{\theta})}$ Factor out $\sin^{2}{\theta}$ in the numerator,
$\frac{\sin^{2}{\theta}(\sin^{2}{\theta} + 1)}{(\sin{\theta} - 1)(1 + \sin{\theta})}$ Expand the denominator,
$\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\sin{\theta} + \sin^{2}{\theta} - 1 - \sin{\theta}}$ Group like terms and simplify in the denominator,
$\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\sin{\theta} - \sin{\theta} + \sin^{2}{\theta} - 1}$ $\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\textcolor{#2192ff}{\sin^{2}{\theta}} - 1}$ Use the identity $\textcolor{#2192ff}{\sin^{2}{\theta} \equiv 1 - \cos^{2}{\theta}}$ in the denominator,
$\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\textcolor{#2192ff}{1 - \cos^{2}{\theta}} - 1}$ Simplify the denominator,
$\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{-\cos^{2}{\theta}}$ Simplify,
$-\tan^{2}{\theta}(1 + \sin^{2}{\theta})$ Therefore,
$\frac{\sin^{3}{\theta}}{\sin{\theta} - 1} - \frac{\sin^{2}{\theta}}{1 + \sin{\theta}} \equiv -\tan^{2}{\theta}(1 + \sin^{2}{\theta})$

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