Ninth Grade Mathematics: Fundamental Identity: sin²θ + cos²θ = 1
Trigonometric identities are mathematical equations that hold true for every possible acute angle theta. The most famous and essential of these is the Fundamental Pythagorean Identity, sin^2(theta) + cos^2(theta) = 1, which translates the Pythagorean theorem directly into the language of angles. In this chapter, you will master the derivation of this master identity and its two companion forms involving secant, tangent, cosecant, and cotangent.
Derivation of the Master Identity
Consider any right-angled triangle ABC with right angle at B, acute angle theta at vertex C, opposite leg AB = O, adjacent leg BC = A, and hypotenuse AC = H.
A
*
/|
/ |
Hypotenuse H / | Opposite O
/ |
/ |
*-----+
C A B
Angle theta
By the Pythagorean theorem:
O^2 + A^2 = H^2
Divide every term on both sides of this equation by H^2 (the square of the hypotenuse):
O^2 A^2 H^2
----- + ----- = -----
H^2 H^2 H^2
(O / H)^2 + (A / H)^2 = 1
By definition of the primary trigonometric ratios, O/H = sin(theta) and A/H = cos(theta). Substituting these yields the Fundamental Pythagorean Identity:
The Fundamental Pythagorean Identity:
sin^2(theta) + cos^2(theta) = 1
(Notation: sin^2(theta) means (sin(theta))^2, NOT sin(theta^2))
This identity holds true for every real angle. If you know the value of sine, cosine is immediately determined by cos(theta) = sqrt(1 - sin^2(theta)), and vice versa.
The Two Companion Pythagorean Identities
By dividing the master identity by cos^2(theta) or sin^2(theta), we generate two equally vital companion identities:
Derivation of Companion Identity 1:
Divide sin^2(theta) + cos^2(theta) = 1 by cos^2(theta):
sin^2(theta) cos^2(theta) 1
-------------- + -------------- = --------------
cos^2(theta) cos^2(theta) cos^2(theta)
tan^2(theta) + 1 = sec^2(theta)
Identity 2:
1 + tan^2(theta) = sec^2(theta) ===> sec^2(theta) - tan^2(theta) = 1
Derivation of Companion Identity 2:
Divide sin^2(theta) + cos^2(theta) = 1 by sin^2(theta):
sin^2(theta) cos^2(theta) 1
-------------- + -------------- = --------------
sin^2(theta) sin^2(theta) sin^2(theta)
1 + cot^2(theta) = csc^2(theta)
Identity 3:
1 + cot^2(theta) = csc^2(theta) ===> csc^2(theta) - cot^2(theta) = 1
Summary of the Three Pythagorean Identities
Identity Group Primary Form Rearranged Difference Form
----------------------------------------------------------------------------------------
Sine & Cosine sin^2(theta) + cos^2(theta) = 1 1 - sin^2 = cos^2, 1 - cos^2 = sin^2
Secant & Tangent 1 + tan^2(theta) = sec^2(theta) sec^2(theta) - tan^2(theta) = 1
Cosecant & Cotangent 1 + cot^2(theta) = csc^2(theta) csc^2(theta) - cot^2(theta) = 1
Notice the factorization of the difference forms using difference of squares: (sec(theta) - tan(theta)) * (sec(theta) + tan(theta)) = 1 This means sec(theta) - tan(theta) and sec(theta) + tan(theta) are exact algebraic reciprocals!
Proving Trigonometric Identities
To prove that a trigonometric identity is true:
1. Start with the more complicated side (usually the Left-Hand Side, LHS).
2. Convert all tangent, cotangent, secant, and cosecant terms into terms of sine and cosine.
3. Combine fractions over common denominators and apply the Pythagorean identities.
4. Simplify until the expression matches the Right-Hand Side (RHS).
Chapter Exercises
Exercise 1: Prove the trigonometric identity: (1 + tan^2(A)) * (1 - sin(A)) * (1 + sin(A)) = 1.
Exercise 2: Prove the identity: sqrt( (1 + sin(A)) / (1 - sin(A)) ) = sec(A) + tan(A).
Exercise 3: If sec(theta) + tan(theta) = p, express the values of sec(theta), tan(theta), and sin(theta) in terms of p.
Solutions and Explanations
Solution 1: Start with the Left-Hand Side (LHS) = (1 + tan^2(A)) * (1 - sin(A)) * (1 + sin(A)). By the second Pythagorean identity, 1 + tan^2(A) = sec^2(A). The second and third terms form a difference of squares: (1 - sin(A)) * (1 + sin(A)) = 1 - sin^2(A). By the master Pythagorean identity, 1 - sin^2(A) = cos^2(A). Substituting these into the LHS: LHS = sec^2(A) * cos^2(A). Since secant and cosine are reciprocal ratios (sec(A) = 1 / cos(A)), sec^2(A) * cos^2(A) = (1 / cos^2(A)) * cos^2(A) = 1. The LHS equals the RHS, completing the proof.
Solution 2: Start with the LHS = sqrt( (1 + sin(A)) / (1 - sin(A)) ). Multiply the numerator and denominator inside the radical by the conjugate of the denominator, which is (1 + sin(A)): LHS = sqrt( [ (1 + sin(A)) * (1 + sin(A)) ] / [ (1 - sin(A)) * (1 + sin(A)) ] ) = sqrt( (1 + sin(A))^2 / (1 - sin^2(A)) ). Using the Pythagorean identity 1 - sin^2(A) = cos^2(A): LHS = sqrt( (1 + sin(A))^2 / cos^2(A) ). Taking the square root of both numerator and denominator eliminates the squares: LHS = (1 + sin(A)) / cos(A). Split the fraction into two separate terms: LHS = 1 / cos(A) + sin(A) / cos(A). Since 1/cos(A) = sec(A) and sin(A)/cos(A) = tan(A), this simplifies to sec(A) + tan(A). This matches the RHS, completing the proof.
Solution 3: We are given sec(theta) + tan(theta) = p (Equation 1). From the Pythagorean identity sec^2(theta) - tan^2(theta) = 1, factor as a difference of squares: (sec(theta) + tan(theta)) * (sec(theta) - tan(theta)) = 1. Substitute Equation 1: p * (sec(theta) - tan(theta)) = 1, which gives sec(theta) - tan(theta) = 1/p (Equation 2). Add Equation 1 and Equation 2: 2 * sec(theta) = p + 1/p = (p^2 + 1) / p, so sec(theta) = (p^2 + 1) / (2p). Subtract Equation 2 from Equation 1: 2 * tan(theta) = p - 1/p = (p^2 - 1) / p, so tan(theta) = (p^2 - 1) / (2p). To find sin(theta), divide tan(theta) by sec(theta): sin(theta) = tan(theta) / sec(theta) = [ (p^2 - 1) / (2p) ] / [ (p^2 + 1) / (2p) ] = (p^2 - 1) / (p^2 + 1).