Trigonometric Identities - Practice Strategy

1. QUOTIENT IDENTITIES

$\tan \theta =\frac{\sin \theta }{\cos \theta }$ $\cot \theta =\frac{\cos \theta }{\sin \theta }$

2. PYTHAGOREAN IDENTITIES

${\sin }^2\theta +{\cos }^2\theta =1$ ${\tan }^2\theta +1={\sec }^2\theta$ $1+{\cot }^2\theta ={\csc }^2\theta$

3. RECIPROCAL IDENTITIES

$\csc \theta =\frac{1}{\sin \theta }$ $\sec \theta =\frac{1}{\cos \theta }$
$\cot \theta =\frac{1}{\tan \theta }$

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STRATEGY: How to tackle identity problems

Step 1: Look at what you’re given and what you need to find
Step 2: Choose the identity that connects them
Step 3: Substitute and solve

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PRACTICE PROBLEMS

Problem 1: Basic substitution

If $\sin \theta =0.6$, find $\cos \theta$ (assuming $\theta$ is in quadrant I).

Solution: Use: ${\sin }^2\theta +{\cos }^2\theta =1$ $(0.6{)}^2+{\cos }^2\theta =1$ $0.36+{\cos }^2\theta =1$ ${\cos }^2\theta =0.64$ $\cos \theta =0.8$ (positive in quadrant I)

Problem 2: Finding other trig functions

If $\tan \theta =\frac{3}{4}$ and $\theta$ is in quadrant I, find $\sin \theta$ and $\cos \theta$.

Solution: Use: ${\tan }^2\theta +1={\sec }^2\theta$ ${\left(\frac{3}{4}\right)}^2+1={\sec }^2\theta$ $\frac{9}{16}+1={\sec }^2\theta$ $\frac{25}{16}={\sec }^2\theta$ $\sec \theta =\frac{5}{4}$

Since $\sec \theta =\frac{1}{\cos \theta }$: $\cos \theta =\frac{4}{5}$

Since $\tan \theta =\frac{\sin \theta }{\cos \theta }$: $\frac{3}{4}=\frac{\sin \theta }{\frac{4}{5}}$ $\sin \theta =\frac{3}{4}\times \frac{4}{5}=\frac{3}{5}$

Problem 3: Simplifying expressions

Simplify: $\frac{\sin \theta }{\cos \theta }\times \cos \theta$

Solution: $\frac{\sin \theta }{\cos \theta }\times \cos \theta =\sin \theta$

Or recognize: $\frac{\sin \theta }{\cos \theta }=\tan \theta$, so $\tan \theta \times \cos \theta =\sin \theta$

Problem 4: Using reciprocal identities

If $\csc \theta =\frac{5}{3}$, find $\sin \theta$.

Solution: Since $\csc \theta =\frac{1}{\sin \theta }$: $\frac{5}{3}=\frac{1}{\sin \theta }$ $\sin \theta =\frac{3}{5}$

Problem 5: Proving an identity

Prove: ${\tan }^2\theta \times {\cos }^2\theta ={\sin }^2\theta$

Solution: Left side: ${\tan }^2\theta \times {\cos }^2\theta$ Substitute $\tan \theta =\frac{\sin \theta }{\cos \theta }$: ${\left(\frac{\sin \theta }{\cos \theta }\right)}^2\times {\cos }^2\theta$ $=\frac{{\sin }^2\theta }{{\cos }^2\theta }\times {\cos }^2\theta$ $={\sin }^2\theta$ ✓

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COMMON EXAM QUESTION TYPES

Type 1: “Find the exact value”

Given one trig function, find another.

Strategy: Use Pythagorean identities to connect $\sin$, $\cos$, and the others.

Type 2: “Simplify the expression”

Reduce a complex trig expression.

Strategy: Convert everything to $\sin$ and $\cos$ using quotient/reciprocal identities.

Type 3: “Prove the identity”

Show that two expressions are equal.

Strategy: Work with the more complex side, substitute identities until you get the simpler side.

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QUICK REFERENCE TRICKS

Converting to sin and cos (most useful approach):

• $\tan \theta \to \frac{\sin \theta }{\cos \theta }$
• $\sec \theta \to \frac{1}{\cos \theta }$
• $\csc \theta \to \frac{1}{\sin \theta }$
• $\cot \theta \to \frac{\cos \theta }{\sin \theta }$

Most common exam identity:

${\sin }^2\theta +{\cos }^2\theta =1$

Can be rearranged as:

• ${\sin }^2\theta =1-{\cos }^2\theta$
• ${\cos }^2\theta =1-{\sin }^2\theta$

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DRILL PROBLEMS (Do these in 15 minutes)

1. If $\cos \theta =\frac{12}{13}$ (quadrant I), find $\sin \theta$ and $\tan \theta$.

2. If $\sec \theta =2$, find $\cos \theta$ and ${\sin }^2\theta$.

3. Simplify: $\sin \theta \times \csc \theta$

4. Simplify: $\frac{\tan \theta }{\sec \theta }$

5. If $\tan \theta =2$, find ${\sec }^2\theta$.

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ANSWERS TO DRILL PROBLEMS

1. $\sin \theta =\frac{5}{13}$, $\tan \theta =\frac{5}{12}$

2. $\cos \theta =\frac{1}{2}$, ${\sin }^2\theta =\frac{3}{4}$

3. $\sin \theta \times \csc \theta =\sin \theta \times \frac{1}{\sin \theta }=1$

4. $\frac{\tan \theta }{\sec \theta }=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}=\sin \theta$

5. ${\sec }^2\theta ={\tan }^2\theta +1=4+1=5$