Pythagorean Trigonometric Identities Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Pythagorean Trigonometric Identities.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

The fundamental identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 and its derived forms: 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta and 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta.

On the unit circle, the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) is always at distance 1 from the origin. By the Pythagorean theorem, x2+y2=1x^2 + y^2 = 1 becomes cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1. This single fact—that sine and cosine are tied to a circle—generates all three Pythagorean identities. Dividing through by cos⁡2θ\cos^2\theta or sin⁡2θ\sin^2\theta produces the other two forms.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 and its two divided-down forms let you swap one trig function for another.

Common stuck point: The procedure for pythagorean trigonometric identities is the easy part; the trap is reading sin⁡2θ\sin^2\theta as sin⁡(θ2)\sin(\theta^2). Asking "Do I have squared trig functions I can collapse to 1 or swap using the unit-circle relation?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Do I have squared trig functions I can collapse to 1 or swap using the unit-circle relation?

Worked Examples

Example 1

easy
If sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and θ\theta is in Quadrant I, find cos⁡(θ)\cos(\theta) using the Pythagorean identity.

Answer

cos⁡(θ)=45\cos(\theta) = \frac{4}{5}

First step

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

Full solution

  1. 2
    Substitute sin⁡(θ)=35\sin(\theta) = \frac{3}{5}: (35)2+cos⁡2(θ)=1\left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1, so 925+cos⁡2(θ)=1\frac{9}{25} + \cos^2(\theta) = 1.
  2. 3
    Solve: cos⁡2(θ)=1−925=1625\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}, so cos⁡(θ)=±45\cos(\theta) = \pm\frac{4}{5}.
  3. 4
    Since θ\theta is in Quadrant I, cos⁡(θ)>0\cos(\theta) > 0, so cos⁡(θ)=45\cos(\theta) = \frac{4}{5}.
The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 directly relates sine and cosine. When you know one and the quadrant, you can find the other. The quadrant determines the sign of the result.

Example 2

medium
Simplify the expression 1−cos⁡2(θ)sin⁡(θ)cos⁡(θ)\frac{1 - \cos^2(\theta)}{\sin(\theta) \cos(\theta)}.

Example 3

medium
Verify sin⁡x(csc⁡x−sin⁡x)=cos⁡2x\sin x(\csc x - \sin x) = \cos^2 x.

Example 4

hard
Verify the identity 1+sin⁡θcos⁡θ+cos⁡θ1+sin⁡θ=2sec⁡θ\dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} = 2\sec\theta.

Example 5

hard
Verify (sin⁡x+cos⁡x)2=1+2sin⁡xcos⁡x(\sin x + \cos x)^2 = 1 + 2\sin x\cos x.

Example 6

challenge
If sin⁡θ+cos⁡θ=22\sin\theta + \cos\theta = \tfrac{\sqrt{2}}{2}, find sin⁡3θ+cos⁡3θ\sin^3\theta + \cos^3\theta.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

medium
Prove that tan⁡2(θ)+1=sec⁡2(θ)\tan^2(\theta) + 1 = \sec^2(\theta).

Example 2

hard
Simplify sec⁡2(θ)−1csc⁡2(θ)−1\frac{\sec^2(\theta) - 1}{\csc^2(\theta) - 1}.

Example 3

easy
State the fundamental Pythagorean identity.

Example 4

easy
If sin⁡θ=35\sin\theta = \frac{3}{5}, find cos⁡2θ\cos^2\theta.

Example 5

easy
Simplify 1−sin⁡2θ1 - \sin^2\theta.

Example 6

easy
State the identity relating tan⁡θ\tan\theta and sec⁡θ\sec\theta.

Example 7

easy
Simplify sin⁡2θ+cos⁡2θ+3\sin^2\theta + \cos^2\theta + 3.

Example 8

easy
Simplify 1−cos⁡2θ1 - \cos^2\theta.

Example 9

easy
State the identity relating cot⁡θ\cot\theta and csc⁡θ\csc\theta.

Example 10

easy
Simplify sec⁡2θ−tan⁡2θ\sec^2\theta - \tan^2\theta.

Example 11

medium
If cos⁡θ=−45\cos\theta = -\frac{4}{5} and θ\theta is in Quadrant II, find sin⁡θ\sin\theta.

Example 12

medium
If tan⁡θ=512\tan\theta = \frac{5}{12} and θ\theta is in Quadrant I, find sec⁡θ\sec\theta.

Example 13

medium
Simplify 1−cos⁡2θsin⁡θ\frac{1 - \cos^2\theta}{\sin\theta} (assume sin⁡θ≠0\sin\theta \neq 0).

Example 14

medium
Verify the identity cos⁡θ1−sin⁡θ=1+sin⁡θcos⁡θ\frac{\cos\theta}{1 - \sin\theta} = \frac{1 + \sin\theta}{\cos\theta} by cross-multiplying.

Example 15

medium
Simplify tan⁡θ⋅cos⁡θ\tan\theta \cdot \cos\theta.

Example 16

medium
If sin⁡θ=23\sin\theta = \frac{2}{3} and θ\theta is in Quadrant II, find tan⁡θ\tan\theta.

Example 17

medium
Simplify sin⁡2θ1+cos⁡θ\frac{\sin^2\theta}{1 + \cos\theta} (assume cos⁡θ≠−1\cos\theta \neq -1).

Example 18

medium
Simplify cos⁡θ⋅csc⁡θ⋅tan⁡θ\cos\theta \cdot \csc\theta \cdot \tan\theta.

Example 19

medium
If sin⁡θ=725\sin\theta = \frac{7}{25} and θ\theta is in Quadrant I, find cos⁡θ\cos\theta.

Example 20

challenge
Prove the identity 11−sin⁡θ+11+sin⁡θ=2sec⁡2θ\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta} = 2\sec^2\theta.

Example 21

challenge
If sec⁡θ−tan⁡θ=13\sec\theta - \tan\theta = \frac{1}{3}, find sec⁡θ+tan⁡θ\sec\theta + \tan\theta.

Example 22

challenge
Express sin⁡4θ+cos⁡4θ\sin^4\theta + \cos^4\theta in terms of sin⁡2θcos⁡2θ\sin^2\theta\cos^2\theta, then simplify.

Example 23

easy
Simplify csc⁡2θ−cot⁡2θ\csc^2\theta - \cot^2\theta.

Example 24

easy
If cos⁡θ=12\cos\theta = \tfrac{1}{2}, find sin⁡2θ\sin^2\theta.

Example 25

easy
Simplify 5(sin⁡2x+cos⁡2x)5(\sin^2 x + \cos^2 x).

Example 26

easy
Simplify csc⁡2x−1\csc^2 x - 1.

Example 27

easy
If sin⁡θ=0\sin\theta = 0, what is cos⁡2θ\cos^2\theta?

Example 28

medium
Simplify tan⁡xsin⁡x+cos⁡x\tan x\sin x + \cos x.

Example 29

medium
If sin⁡θ=−817\sin\theta = -\tfrac{8}{17} and θ\theta is in Quadrant III, find cos⁡θ\cos\theta.

Example 30

medium
If sec⁡θ=53\sec\theta = \tfrac{5}{3} and θ\theta is in Quadrant IV, find tan⁡θ\tan\theta.

Example 31

medium
Simplify cos⁡2x−1cos⁡x−1\dfrac{\cos^2 x - 1}{\cos x - 1}.

Example 32

medium
Simplify tan⁡xsec⁡x\dfrac{\tan x}{\sec x}.

Example 33

medium
If tan⁡θ=−34\tan\theta = -\tfrac{3}{4} and θ\theta is in Quadrant II, find sec⁡θ\sec\theta and sin⁡θ\sin\theta.

Example 34

medium
Simplify sin⁡2xsec⁡2x+sin⁡2x\sin^2 x\sec^2 x + \sin^2 x.

Example 35

medium
Simplify sin⁡xcot⁡x\sin x\cot x.

Example 36

hard
Solve 2cos⁡2x−1=sin⁡x2\cos^2 x - 1 = \sin x on [0,2π)[0, 2\pi).

Example 37

hard
If sin⁡θ+csc⁡θ=3\sin\theta + \csc\theta = 3, find sin⁡2θ+csc⁡2θ\sin^2\theta + \csc^2\theta.

Example 38

hard
Simplify (sec⁡x−tan⁡x)(sec⁡x+tan⁡x)(\sec x - \tan x)(\sec x + \tan x).

Example 39

hard
Simplify sin⁡6x+cos⁡6x\sin^6 x + \cos^6 x in terms of sin⁡2xcos⁡2x\sin^2 x\cos^2 x.

Example 40

challenge
Find all θ∈[0,2π)\theta \in [0, 2\pi) satisfying sec⁡2θ+tan⁡2θ=3\sec^2\theta + \tan^2\theta = 3.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

trigonometric functionspythagorean theoremunit circle