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Trigonometric Identities

Fundamental trigonometric identities including Pythagorean, reciprocal, quotient, and even-odd identities

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Trigonometric Identities

What Are Identities?

Trigonometric identities are equations that are true for all values of the variable (where both sides are defined).

They are powerful tools for simplifying expressions and solving equations.

Reciprocal Identities

csc⁡(θ)=1sin⁡(θ)\csc(\theta) = \frac{1}{\sin(\theta)}

sec⁡(θ)=1cos⁡(θ)\sec(\theta) = \frac{1}{\cos(\theta)}

cot⁡(θ)=1tan⁡(θ)\cot(\theta) = \frac{1}{\tan(\theta)}

Quotient Identities

tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

cot⁡(θ)=cos⁡(θ)sin⁡(θ)\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}

Pythagorean Identities

These come from x2+y2=1x^2 + y^2 = 1 on the unit circle:

Primary Form

sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

Divide by cos⁡2(θ)\cos^2(\theta)

tan⁡2(θ)+1=sec⁡2(θ)\tan^2(\theta) + 1 = \sec^2(\theta)

Divide by sin⁡2(θ)\sin^2(\theta)

1+cot⁡2(θ)=csc⁡2(θ)1 + \cot^2(\theta) = \csc^2(\theta)

Even-Odd Identities

Even functions (symmetric about y-axis): cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta) sec⁡(−θ)=sec⁡(θ)\sec(-\theta) = \sec(\theta)

Odd functions (symmetric about origin): sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta) tan⁡(−θ)=−tan⁡(θ)\tan(-\theta) = -\tan(\theta) csc⁡(−θ)=−csc⁡(θ)\csc(-\theta) = -\csc(\theta) cot⁡(−θ)=−cot⁡(θ)\cot(-\theta) = -\cot(\theta)

Cofunction Identities

Cofunctions of complementary angles are equal:

sin⁡(π2−θ)=cos⁡(θ)\sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta)

cos⁡(π2−θ)=sin⁡(θ)\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)

tan⁡(π2−θ)=cot⁡(θ)\tan\left(\frac{\pi}{2} - \theta\right) = \cot(\theta)

How to Use Identities

  1. Simplify expressions: Replace complex trig expressions with simpler ones
  2. Prove other identities: Use known identities to verify new ones
  3. Solve equations: Transform equations into solvable forms
  4. Evaluate expressions: Find exact values

Strategy for Proving Identities

  1. Start with the more complicated side
  2. Use fundamental identities to rewrite terms
  3. Look for opportunities to factor or combine fractions
  4. Convert everything to sines and cosines if stuck
  5. Never move terms from one side to the other (work on each side independently)

Common Mistakes to Avoid

❌ Don't treat identities like equations and cross-multiply ❌ Don't forget to square correctly: (sin⁡θ)2=sin⁡2θ(\sin \theta)^2 = \sin^2 \theta ❌ Don't cancel terms that aren't factors

📚 Practice Problems

1Problem 1medium

❓ Question:

Simplify the expression: sin⁡(θ)cos⁡(θ)+cos⁡(θ)sin⁡(θ)\frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}

💡 Show Solution

Solution:

Step 1: Recognize the quotient identities. sin⁡(θ)cos⁡(θ)=tan⁡(θ)\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) cos⁡(θ)sin⁡(θ)=cot⁡(θ)\frac{\cos(\theta)}{\sin(\theta)} = \cot(\theta)

So our expression becomes: tan⁡(θ)+cot⁡(θ)\tan(\theta) + \cot(\theta)

Step 2: Find a common denominator. =sin⁡(θ)cos⁡(θ)+cos⁡(θ)sin⁡(θ)= \frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}

=sin⁡2(θ)+cos⁡2(θ)sin⁡(θ)cos⁡(θ)= \frac{\sin^2(\theta) + \cos^2(\theta)}{\sin(\theta)\cos(\theta)}

Step 3: Use the Pythagorean identity. sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

=1sin⁡(θ)cos⁡(θ)= \frac{1}{\sin(\theta)\cos(\theta)}

Step 4: Write in terms of reciprocal identities. =1sin⁡(θ)⋅1cos⁡(θ)=csc⁡(θ)sec⁡(θ)= \frac{1}{\sin(\theta)} \cdot \frac{1}{\cos(\theta)} = \csc(\theta)\sec(\theta)

Answer: csc⁡(θ)sec⁡(θ)\csc(\theta)\sec(\theta) or 1sin⁡(θ)cos⁡(θ)\frac{1}{\sin(\theta)\cos(\theta)}

2Problem 2medium

❓ Question:

Verify the following identities:

a) tan⁡xcos⁡x=sin⁡x\tan x \cos x = \sin x b) 1−sin⁡2xcos⁡x=cos⁡x\frac{1 - \sin^2 x}{\cos x} = \cos x c) sec⁡2x−1=tan⁡2x\sec^2 x - 1 = \tan^2 x

💡 Show Solution

Solution:

Part (a): Start with the left side:

tan⁡xcos⁡x=sin⁡xcos⁡x⋅cos⁡x=sin⁡x\tan x \cos x = \frac{\sin x}{\cos x} \cdot \cos x = \sin x ✓

Part (b): Start with the left side:

1−sin⁡2xcos⁡x\frac{1 - \sin^2 x}{\cos x}

Use Pythagorean identity: sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, so 1−sin⁡2x=cos⁡2x1 - \sin^2 x = \cos^2 x

=cos⁡2xcos⁡x=cos⁡x= \frac{\cos^2 x}{\cos x} = \cos x ✓

Part (c): Start with the left side:

sec⁡2x−1\sec^2 x - 1

Recall: sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}

We can use the Pythagorean identity: tan⁡2x+1=sec⁡2x\tan^2 x + 1 = \sec^2 x

Rearranging: sec⁡2x−1=tan⁡2x\sec^2 x - 1 = \tan^2 x ✓

Alternatively, we could derive it:

sec⁡2x−1=1cos⁡2x−1=1−cos⁡2xcos⁡2x=sin⁡2xcos⁡2x=tan⁡2x\sec^2 x - 1 = \frac{1}{\cos^2 x} - 1 = \frac{1 - \cos^2 x}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x ✓

3Problem 3easy

❓ Question:

Prove the identity: 1−sin⁡2(x)cos⁡(x)=cos⁡(x)\frac{1 - \sin^2(x)}{\cos(x)} = \cos(x)

💡 Show Solution

Solution:

We'll work with the left side to show it equals the right side.

Left side:

Step 1: Recognize the Pythagorean identity.

We know: sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1

Therefore: 1−sin⁡2(x)=cos⁡2(x)1 - \sin^2(x) = \cos^2(x)

Step 2: Substitute.

1−sin⁡2(x)cos⁡(x)=cos⁡2(x)cos⁡(x)\frac{1 - \sin^2(x)}{\cos(x)} = \frac{\cos^2(x)}{\cos(x)}

Step 3: Simplify.

=cos⁡(x)⋅cos⁡(x)cos⁡(x)=cos⁡(x)= \frac{\cos(x) \cdot \cos(x)}{\cos(x)} = \cos(x)

This equals the right side! ✓

The identity is proven.

4Problem 4hard

❓ Question:

Simplify the expression:

sin⁡x1+cos⁡x+1+cos⁡xsin⁡x\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x}

💡 Show Solution

Solution:

Find a common denominator:

sin⁡x1+cos⁡x+1+cos⁡xsin⁡x=sin⁡2x+(1+cos⁡x)2(1+cos⁡x)sin⁡x\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = \frac{\sin^2 x + (1 + \cos x)^2}{(1 + \cos x)\sin x}

Expand the numerator:

sin⁡2x+(1+cos⁡x)2=sin⁡2x+1+2cos⁡x+cos⁡2x\sin^2 x + (1 + \cos x)^2 = \sin^2 x + 1 + 2\cos x + \cos^2 x

=(sin⁡2x+cos⁡2x)+1+2cos⁡x= (\sin^2 x + \cos^2 x) + 1 + 2\cos x

=1+1+2cos⁡x= 1 + 1 + 2\cos x (using sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1)

=2+2cos⁡x=2(1+cos⁡x)= 2 + 2\cos x = 2(1 + \cos x)

So our expression becomes:

2(1+cos⁡x)(1+cos⁡x)sin⁡x=2sin⁡x=2csc⁡x\frac{2(1 + \cos x)}{(1 + \cos x)\sin x} = \frac{2}{\sin x} = 2\csc x

Final answer: 2csc⁡x2\csc x

5Problem 5medium

❓ Question:

If sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and θ\theta is in Quadrant II, find cos⁡(θ)\cos(\theta) and tan⁡(θ)\tan(\theta).

💡 Show Solution

Solution:

Given: sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and θ\theta is in Quadrant II

Find cos⁡(θ)\cos(\theta):

Step 1: Use the Pythagorean identity. sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

(35)2+cos⁡2(θ)=1\left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1

925+cos⁡2(θ)=1\frac{9}{25} + \cos^2(\theta) = 1

cos⁡2(θ)=1−925=1625\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}

cos⁡(θ)=±45\cos(\theta) = \pm\frac{4}{5}

Step 2: Determine the sign.

In Quadrant II, cosine is negative.

cos⁡(θ)=−45\cos(\theta) = -\frac{4}{5}

Find tan⁡(θ)\tan(\theta):

Use the quotient identity: tan⁡(θ)=sin⁡(θ)cos⁡(θ)=3/5−4/5=35⋅5−4=−34\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{3/5}{-4/5} = \frac{3}{5} \cdot \frac{5}{-4} = -\frac{3}{4}

Answers:

  • cos⁡(θ)=−45\cos(\theta) = -\frac{4}{5}
  • tan⁡(θ)=−34\tan(\theta) = -\frac{3}{4}
Explain using:

⚠️ Common Mistakes: Trigonometric Identities

Avoid these 4 frequent errors

🌍 Real-World Applications: Trigonometric Identities

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values
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📌 Related Topics in Trigonometric Functions

Degrees and RadiansUnit Circle and Radian Measure (Combined - See Split Topics)Arc Length and Sector AreaThe Unit CircleNegative and Coterminal AnglesSum and Difference Identities

❓ Frequently Asked Questions

What is Trigonometric Identities?▾
Fundamental trigonometric identities including Pythagorean, reciprocal, quotient, and even-odd identities
How can I study Trigonometric Identities effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Trigonometric Identities study guide free?▾
Yes — all study notes, flashcards, and practice problems for Trigonometric Identities on Study Mondo are free to access. No account is needed.
What course covers Trigonometric Identities?▾
Trigonometric Identities is part of the AP Precalculus course on Study Mondo, specifically in the Trigonometric Functions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Trigonometric Identities?▾
Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.