Equivalent Representations of Trigonometric Functions

Worked example

1. 01

   Rewrite $\sec \theta = 1/\cos \theta$. The numerator is $1/\cos \theta - \cos \theta = (1 - \cos ^{2} \theta )/\cos \theta$.

2. 02

   Pythagorean identity: $1 - \cos ^{2} \theta = \sin ^{2} \theta$. So the numerator is $\sin ^{2} \theta / \cos \theta$.

3. 03

   Divide by $\sin \theta$: $(\sin ^{2} \theta / \cos \theta ) \div \sin \theta = \sin \theta / \cos \theta = \tan \theta$.

Worked example

1. 01

   Replace each factor by a Pythagorean identity: $1 - \cos ^{2} \theta = \sin ^{2} \theta$ and $1 + \cot ^{2} \theta = \csc ^{2} \theta$.

2. 02

   $\sin ^{2} \theta \cdot \csc ^{2} \theta = \sin ^{2} \theta \cdot (1/\sin ^{2} \theta ) = 1$. The whole expression equals 1 (wherever it is defined).

   To prove an identity, start with ONE side and transform it step by step until it becomes the other side. Do not treat it like an equation to solve, and do not do the same operation to both sides — that assumes what you are trying to prove.

Worked example

1. 01

   Start with the left side, which has more to simplify: $\csc \theta - \sin \theta = 1/\sin \theta - \sin \theta = (1 - \sin ^{2} \theta )/\sin \theta$.

2. 02

   Pythagorean identity: $1 - \sin ^{2} \theta = \cos ^{2} \theta$, so the left side is $\cos ^{2} \theta / \sin \theta$.

3. 03

   Split the product: $\cos ^{2} \theta / \sin \theta = \cos \theta \cdot (\cos \theta / \sin \theta ) = \cos \theta \cot \theta$, which is the right side ✓.

Worked example

1. 01

   Two functions appear. Replace $\cos ^{2} \theta$ with $1 - \sin ^{2} \theta$: $2(1 - \sin ^{2} \theta ) + 3 \sin \theta = 3$.

2. 02

   Expand and collect on one side: $2 \sin ^{2} \theta - 3 \sin \theta + 1 = 0$, which factors as $(2 \sin \theta - 1)(\sin \theta - 1) = 0$.

3. 03

   $\sin \theta = 1/2$ gives $\theta = \pi /6$ and $5\pi /6$. $\sin \theta = 1$ gives $\theta = \pi /2$.

   Graphical check: $y = 2\cos ^{2}\theta + 3 \sin \theta$ meets $y = 3$ at $\pi /6$, $\pi /2$, and $5\pi /6$ (it just touches at $\pi /2$).

Worked example

1. 01

   $\tan ^{2} \theta$ and $\sec \theta$ are linked by $1 + \tan ^{2} \theta = \sec ^{2} \theta$. Replace $\tan ^{2} \theta$ with $\sec ^{2} \theta - 1$: $\sec ^{2} \theta - 1 = 2 \sec \theta + 2$.

2. 02

   Collect: $\sec ^{2} \theta - 2 \sec \theta - 3 = 0$, which factors as $(\sec \theta - 3)(\sec \theta + 1) = 0$.

3. 03

   $\sec \theta = 3$ means $\cos \theta = 1/3$: $\theta \approx 1.231$ and $2\pi - 1.231 \approx 5.052$ (the same values as the mistake box in 3.10A). $\sec \theta = -1$ means $\cos \theta = -1$: $\theta = \pi$.

4. 04

   Check that nothing is undefined: $\tan \theta$ and $\sec \theta$ exist at all three angles, since $\cos \theta \ne 0$ there ✓.