Prove That sin²θ + cos²θ = 1 | Class 10 Maths Ch 8 Q 4
This is one of the most fundamental and important trigonometric identities you’ll encounter in Class 10 Mathematics. The Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\) forms the backbone of trigonometry and appears in countless problems throughout your board exams. Understanding this proof will not only help you score full marks on this question but also strengthen your grasp of the relationship between sine and cosine functions.
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Proof: In a right-angled triangle with angle θ, by Pythagoras theorem: \(\text{(Perpendicular)}^2 + \text{(Base)}^2 = \text{(Hypotenuse)}^2\). Dividing throughout by \(\text{(Hypotenuse)}^2\), we get \(\left(\frac{\text{Perpendicular}}{\text{Hypotenuse}}\right)^2 + \left(\frac{\text{Base}}{\text{Hypotenuse}}\right)^2 = 1\). Since \(\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\) and \(\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}}\), we get \(\sin^2\theta + \cos^2\theta = 1\).
Understanding the Pythagorean Trigonometric Identity
The identity \(\sin^2\theta + \cos^2\theta = 1\) is called the Pythagorean identity because it’s directly derived from the Pythagorean theorem. This relationship holds true for any angle θ, making it one of the most versatile tools in trigonometry.
Why This Identity is Important
- Foundation of Trigonometry: This is the most fundamental relationship between sine and cosine
- Problem Solving: Used to simplify complex trigonometric expressions
- Exam Frequency: Appears in 80% of trigonometry questions, either directly or indirectly
- Practical Applications: Used in physics, engineering, and computer graphics
- Derivation Base: Other identities like \(1 + \tan^2\theta = \sec^2\theta\) are derived from this
Prerequisites for Understanding
Before we dive into the proof, make sure you’re comfortable with:
- Pythagoras Theorem: In a right triangle, \(a^2 + b^2 = c^2\)
- Sine Definition: \(\sin\theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}\)
- Cosine Definition: \(\cos\theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}\)
- Basic Algebra: Dividing equations and simplifying fractions
📐 Visual Representation
Right-Angled Triangle ABC:
C (Right angle)
|\
| \
| \ c (Hypotenuse)
a | \
| \
| \
|θ \
A-------B
b
Where:
• Angle at A = θ
• Angle at C = 90°
• Side BC (opposite to θ) = a (Perpendicular)
• Side AC (adjacent to θ) = b (Base)
• Side AB (hypotenuse) = c
By Pythagoras Theorem:
a² + b² = c²
Trigonometric Ratios:
sin θ = a/c (Perpendicular/Hypotenuse)
cos θ = b/c (Base/Hypotenuse)
🧠 Memory Trick
“Sine Squared Plus Cosine Squared is Always ONE!”
Remember: S²+C²=1 (Sine squared plus Cosine squared equals one)
Think of it as: “Sin and Cos are best friends who together make a complete unit (1)”
Visual: Imagine a circle with radius 1 – the horizontal and vertical components always add up to 1 when squared!
📝 How to Write This Answer in Your Exam (3 Marks)
📊 Marking Scheme Breakdown:
- 1 Mark: Drawing the right triangle and applying Pythagoras theorem
- 1 Mark: Dividing by hypotenuse² and simplifying
- 1 Mark: Substituting sine and cosine definitions to reach the final identity
✍️ Perfect Exam Answer:
To Prove: \(\sin^2\theta + \cos^2\theta = 1\)
Proof:
Consider a right-angled triangle ABC, right-angled at C, with angle A = θ.
Let BC = a (perpendicular), AC = b (base), and AB = c (hypotenuse).
By Pythagoras theorem:
\[a^2 + b^2 = c^2\]
Dividing both sides by \(c^2\):
\[\frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{c^2}{c^2}\]
\[\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1\]
Since \(\sin\theta = \frac{a}{c}\) and \(\cos\theta = \frac{b}{c}\), we have:
\[\sin^2\theta + \cos^2\theta = 1\]
Hence proved.
⭐ Exam Tip: Always write “To Prove” and “Hence Proved” to show complete structure. Draw a neat diagram if time permits – it can earn you presentation marks!
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Forgetting to square the ratios
Students often write \(\sin\theta + \cos\theta = 1\) instead of \(\sin^2\theta + \cos^2\theta = 1\). Remember, both terms must be squared!
❌ Mistake 2: Wrong division step
When dividing \(a^2 + b^2 = c^2\) by \(c^2\), some students forget to divide all three terms. Make sure you divide every term by \(c^2\).
❌ Mistake 3: Confusion with other identities
Don’t mix this up with \(1 + \tan^2\theta = \sec^2\theta\) or \(1 + \cot^2\theta = \csc^2\theta\). Each identity has its own specific form.
✅ Complete Step-by-Step Proof
Step 1: Set Up the Right Triangle
Consider a right-angled triangle ABC where:
- Angle C = 90° (right angle)
- Angle A = θ (the angle we’re working with)
- Side BC (opposite to θ) = a
- Side AC (adjacent to θ) = b
- Side AB (hypotenuse) = c
Step 2: Apply Pythagoras Theorem
In any right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides:
\[a^2 + b^2 = c^2\]
This is our starting point. Everything else follows from this fundamental theorem.
Step 3: Divide by c² (Hypotenuse Squared)
Now, divide every term of the equation by \(c^2\):
\[\frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{c^2}{c^2}\]
Why divide by c²? Because we want to create ratios that match the definitions of sine and cosine.
Step 4: Simplify the Fractions
We can rewrite each fraction as a squared term:
\[\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1\]
Notice how \(\frac{c^2}{c^2} = 1\) on the right side. The left side now has two perfect squared ratios.
Step 5: Substitute Trigonometric Definitions
Recall the definitions:
- \(\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a}{c}\)
- \(\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{b}{c}\)
Substituting these into our equation:
\[(\sin\theta)^2 + (\cos\theta)^2 = 1\]
\[\sin^2\theta + \cos^2\theta = 1\]
🎉 Proof Complete!
\(\sin^2\theta + \cos^2\theta = 1\)
This identity is true for all values of θ
🔄 Alternative Way to Think About It
Unit Circle Approach
If you’re comfortable with the unit circle concept, here’s another perspective:
- Consider a circle with radius 1 (unit circle)
- Any point on this circle has coordinates \((\cos\theta, \sin\theta)\)
- The distance from origin to any point on the circle is always 1
- Using distance formula: \(\sqrt{(\cos\theta)^2 + (\sin\theta)^2} = 1\)
- Squaring both sides: \(\cos^2\theta + \sin^2\theta = 1\)
This geometric interpretation shows why the identity works for all angles, not just acute angles in a right triangle!
💡 Where This Identity is Used
📐 Simplifying Expressions
Replace \(\sin^2\theta\) with \(1 – \cos^2\theta\) or vice versa to simplify complex trigonometric expressions.
🔢 Solving Equations
Convert equations with both sine and cosine into equations with just one function.
⚡ Physics Problems
Used in wave motion, oscillations, and resolving vectors into components.
🎮 Computer Graphics
Rotations, transformations, and 3D modeling all rely on this fundamental identity.
❓ Frequently Asked Questions
✍️ Written by Farhan Mansuri
Executive Officer, BSNL
Bharat Sanchar Nigam Limited • Khargone, Madhya Pradesh, India
Farhan Mansuri is an Executive Officer at Bharat Sanchar Nigam Limited with over 25 years of combined experience in telecommunications and education. He specializes in making complex mathematical concepts accessible to students and has helped thousands of learners excel in their board examinations. His practical approach combines real-world applications with exam-focused strategies, ensuring students not only understand the concepts but also know exactly how to present them in exams for maximum marks.
📚 Related Questions You Might Find Helpful
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- Simplify: (sin²θ – cos²θ) + 1
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