Trigonometric Identities

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

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}
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