1.5.3 Trigonometric Identities

In this topic we will learn how to:

  • use the identities sin⁡θcos⁡θ≡tan⁡θ\frac{\sin{\theta}}{\cos{\theta}} \equiv \tan{\theta} and sin⁡2θ+cos⁡2θ≡1\sin^{2}{\theta} + \cos^{2}{\theta} \equiv 1

Pythagorean Identities\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⁡θ≡sin⁡θcos⁡θ\tan{\theta} \equiv \frac{\sin{\theta}}{\cos{\theta}}sin⁡2θ+cos⁡2θ≡1\sin^{2}{\theta} + \cos^{2}{\theta} \equiv 1In some cases, the second trig identity, may be useful in the forms below,
sin⁡2θ≡1−cos⁡2θ\sin^{2}{\theta} \equiv 1 - \cos^{2}{\theta}cos⁡2θ≡1−sin⁡2θ\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\tan{x} to be in terms of sin⁡x\sin{x} and cos⁡x\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,

sin⁡θ+2cos⁡θcos⁡θ−2sin⁡θ−sin⁡θ−2cos⁡θcos⁡θ+2sin⁡θ≡45cos⁡2θ−4\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,
sin⁡θ+2cos⁡θcos⁡θ−2sin⁡θ−sin⁡θ−2cos⁡θcos⁡θ+2sin⁡θ\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,

(sin⁡θ+2cos⁡θ)(cos⁡θ+2sin⁡θ)−(sin⁡θ−2cos⁡θ)(cos⁡θ−2sin⁡θ)(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\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,

sin⁡θcos⁡θ+2sin⁡2θ+2cos⁡2θ+4sin⁡θcos⁡θ−(−2sin⁡2θ+sin⁡θcos⁡θ−2cos⁡2θ+4sin⁡θcos⁡θ)(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\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,

5sin⁡θcos⁡θ+2sin⁡2θ+2cos⁡2θ−(−2sin⁡2θ−2cos⁡2θ+5sin⁡θcos⁡θ)(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\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})}

5sin⁡θcos⁡θ+2(sin⁡2θ+cos⁡2θ)−(−2(sin⁡2θ+cos⁡2θ)+5sin⁡θcos⁡θ)(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\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 sin⁡2θ+cos⁡2θ≡1\textcolor{#2192ff}{\sin^{2}{\theta} + \cos^{2}{\theta} \equiv 1},
5sin⁡θcos⁡θ+2(1)−(−2(1)+5sin⁡θcos⁡θ)(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\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})}5sin⁡θcos⁡θ+2−(−2+5sin⁡θcos⁡θ)(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\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,
5sin⁡θcos⁡θ+2+2−5sin⁡θcos⁡θ(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\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,
4(cos⁡θ−2sin⁡θ)(cos⁡θ+2sin⁡θ)\frac{4}{(\cos{\theta} - 2\sin{\theta})(\cos{\theta} + 2\sin{\theta})}Expand the denominator,
4cos⁡2θ+2sin⁡θcos⁡θ−2sin⁡θcos⁡θ−4sin⁡2θ\frac{4}{\cos^{2}{\theta} + 2\sin{\theta} \cos{\theta} - 2\sin{\theta} \cos{\theta} - 4\sin^{2}{\theta}}Simplify the denominator,
4cos⁡2θ−4sin⁡2θ\frac{4}{\cos^{2}{\theta} - 4\textcolor{#2192ff}{\sin^{2}{\theta}}}Use the identity sin⁡2x≡1−cos⁡2x\textcolor{#2192ff}{\sin^{2}{x} \equiv 1 - \cos^{2}{x}} in the denominator,
4cos⁡2θ−4(1−cos⁡2θ)\frac{4}{\cos^{2}{\theta} - 4(\textcolor{#2192ff}{1 - \cos^{2}{\theta}})}Simplify the denominator,
4cos⁡2θ−4+4cos⁡2θ\frac{4}{\cos^{2}{\theta} - 4 + 4\cos^{2}{\theta}}Group like terms in the denominator,4cos⁡2θ+4cos⁡2θ−4\frac{4}{\cos^{2}{\theta} + 4\cos^{2}{\theta} - 4}45cos⁡2θ−4\frac{4}{5\cos^{2}{\theta} - 4}Therefore,
sin⁡θ+2cos⁡θcos⁡θ−2sin⁡θ−sin⁡θ−2cos⁡θcos⁡θ+2sin⁡θ≡45cos⁡2θ−4\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,

1+sin⁡x1−sin⁡x−1−sin⁡x1+sin⁡x≡4tan⁡xcos⁡x\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,
1+sin⁡x1−sin⁡x−1−sin⁡x1+sin⁡x\frac{1 + \sin{x}}{1 - \sin{x}} - \frac{1 - \sin{x}}{1 + \sin{x}}Compose the two fractions into one fraction,
(1+sin⁡x)2−(1−sin⁡x)2(1−sin⁡x)(1+sin⁡x)\frac{(1 + \sin{x})^{2} - (1 - \sin{x})^{2}}{(1 - \sin{x})(1 + \sin{x})}Expand the numerator,
1+2sin⁡x+sin⁡2x−(1−2sin⁡x+sin⁡2x)(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})}1+2sin⁡x+sin⁡2x−1+2sin⁡x−sin⁡2x(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,
1−1+2sin⁡x+2sin⁡x+sin⁡2x−sin⁡2x(1−sin⁡x)(1+sin⁡x)\frac{1 - 1 + 2\sin{x} + 2\sin{x} + \sin^{2}{x} - \sin^{2}{x}}{(1 - \sin{x})(1 + \sin{x})}4sin⁡x(1−sin⁡x)(1+sin⁡x)\frac{4\sin{x}}{(1 - \sin{x})(1 + \sin{x})}Expand the denominator,
4sin⁡x1−sin⁡2x\frac{4\sin{x}}{\textcolor{#2192ff}{1 - \sin^{2}{x}}}Use the identity cos⁡2x≡1−sin⁡2x\textcolor{#2192ff}{\cos^{2}{x} \equiv 1 - \sin^{2}{x}} in the denominator,
4sin⁡xcos⁡2x\frac{4\sin{x}}{\textcolor{#2192ff}{\cos^{2}{x}}}Simplify,
4sin⁡xcos⁡x×cos⁡x\frac{4\sin{x}}{\cos{x} \times \cos{x}}4sin⁡xcos⁡x×1cos⁡x\frac{4\sin{x}}{\cos{x}} \times \frac{1}{\cos{x}}4tan⁡x×1cos⁡x4\tan{x} \times \frac{1}{\cos{x}}4tan⁡xcos⁡x\frac{4\tan{x}}{\cos{x}}Therefore,
1+sin⁡x1−sin⁡x−1−sin⁡x1+sin⁡x≡4tan⁡xcos⁡x\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,

1−2sin⁡2θ1−sin⁡2θ=1−tan⁡2θ\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−tan⁡2θ1 - \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⁡2θ\tan^{2}{\theta} to be in terms of sin⁡θ\sin{\theta} and cos⁡θ\cos{\theta}.

Use the identity tan⁡θ=sin⁡θcos⁡θ\textcolor{#2192ff}{\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}},
1−sin⁡2θcos⁡2θ1 - \textcolor{#2192ff}{\frac{\sin^{2}{\theta}}{\cos^{2}{\theta}}}Compose the two terms into one fraction,
cos⁡2θ−sin⁡2θcos⁡2θ\frac{\textcolor{#0f0}{\cos^{2}{\theta}} - \sin^{2}{\theta}}{\cos^{2}{\theta}}Use the identity cos⁡2≡1−sin⁡2θ\textcolor{#0f0}{\cos^{2} \equiv 1 - \sin^{2}{\theta}} in the numerator,
1−sin⁡2θ−sin⁡2θcos⁡2θ\frac{\textcolor{#0f0}{1 - \sin^{2}{\theta}} - \sin^{2}{\theta}}{\cos^{2}{\theta}}Simplify the numerator,
1−2sin⁡2θcos⁡2θ\frac{1 - 2\sin^{2}{\theta}}{\textcolor{red}{\cos^{2}{\theta}}}Use the identity cos⁡2≡1−sin⁡2θ\textcolor{red}{\cos^{2} \equiv 1 - \sin^{2}{\theta}} in the denominator,
1−2sin⁡2θ1−sin⁡2θ\frac{1 - 2\sin^{2}{\theta}}{\textcolor{red}{1 - \sin^{2}{\theta}}}Therefore,
1−2sin⁡2θ1−sin⁡2θ=1−tan⁡2θ\frac{1 - 2\sin^{2}{\theta}}{1 - \sin^{2}{\theta}} = 1 - \tan^{2}{\theta}4. Prove the identity,

sin⁡3θsin⁡θ−1−sin⁡2θ1+sin⁡θ≡−tan⁡2θ(1+sin⁡2θ)\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,
sin⁡3θsin⁡θ−1−sin⁡2θ1+sin⁡θ\frac{\sin^{3}{\theta}}{\sin{\theta} - 1} - \frac{\sin^{2}{\theta}}{1 + \sin{\theta}}Compose the two fractions into one fraction,
sin⁡3θ(1+sin⁡θ)−sin⁡2θ(sin⁡θ−1)(sin⁡θ−1)(1+sin⁡θ)\frac{\sin^{3}{\theta}(1 + \sin{\theta}) - \sin^{2}{\theta}(\sin{\theta} - 1)}{(\sin{\theta} - 1)(1 + \sin{\theta})}Expand the numerator and simplify,
sin⁡3θ+sin⁡4θ−sin⁡3θ+sin⁡2θ(sin⁡θ−1)(1+sin⁡θ)\frac{\sin^{3}{\theta} + \sin^{4}{\theta} - \sin^{3}{\theta} + \sin^{2}{\theta}}{(\sin{\theta} - 1)(1 + \sin{\theta})}sin⁡4θ+sin⁡2θ(sin⁡θ−1)(1+sin⁡θ)\frac{\sin^{4}{\theta} + \sin^{2}{\theta}}{(\sin{\theta} - 1)(1 + \sin{\theta})}Factor out sin⁡2θ\sin^{2}{\theta} in the numerator,
sin⁡2θ(sin⁡2θ+1)(sin⁡θ−1)(1+sin⁡θ)\frac{\sin^{2}{\theta}(\sin^{2}{\theta} + 1)}{(\sin{\theta} - 1)(1 + \sin{\theta})}Expand the denominator,
sin⁡2θ(1+sin⁡2θ)sin⁡θ+sin⁡2θ−1−sin⁡θ\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\sin{\theta} + \sin^{2}{\theta} - 1 - \sin{\theta}}Group like terms and simplify in the denominator,
sin⁡2θ(1+sin⁡2θ)sin⁡θ−sin⁡θ+sin⁡2θ−1\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\sin{\theta} - \sin{\theta} + \sin^{2}{\theta} - 1}sin⁡2θ(1+sin⁡2θ)sin⁡2θ−1\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\textcolor{#2192ff}{\sin^{2}{\theta}} - 1}Use the identity sin⁡2θ≡1−cos⁡2θ\textcolor{#2192ff}{\sin^{2}{\theta} \equiv 1 - \cos^{2}{\theta}} in the denominator,
sin⁡2θ(1+sin⁡2θ)1−cos⁡2θ−1\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{\textcolor{#2192ff}{1 - \cos^{2}{\theta}} - 1}Simplify the denominator,
sin⁡2θ(1+sin⁡2θ)−cos⁡2θ\frac{\sin^{2}{\theta}(1 + \sin^{2}{\theta})}{-\cos^{2}{\theta}}Simplify,
−tan⁡2θ(1+sin⁡2θ)-\tan^{2}{\theta}(1 + \sin^{2}{\theta})Therefore,
sin⁡3θsin⁡θ−1−sin⁡2θ1+sin⁡θ≡−tan⁡2θ(1+sin⁡2θ)\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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