Equivalent Representations of Trigonometric Functions

Equivalent representations of trigonometric functions refer to different ways of expressing the functions. They are done using trigonometric identities and transformations (phase shifts). For example， cot x sin2 x = 3 cot x ⇒ cot x sin2 x 3 cot x = 0 ⇒ cot x(sin2 x 3) = 0 Either cot x = 0 or sin2 x 3 = 0 If cot x = 0， they repeat every 2π and thus are related as follows: sin θ = sin(2nπ + θ) cos θ = cos (2nπ + θ) tan θ = tan (nπ + θ) cot θ = cot (nπ + θ) sec θ = sec (2nπ + θ) cosec θ = cosec (2nπ + θ) Solved Examples What is the value of sin(3x + 45)° if sin x = 0.5 and cos x  = 0.866 Solution: As we know。

the triple angles identities:  sin 3θ = 3 sin θ 4 sin3 θ  cos 3θ = 4 cos3 θ 3 cos θ Given sin x = 0.5 and cos x = 0.866 From (i)， graphing functions， x = ${\dfrac{\pi }{2}+n\pi}$， \theta =\dfrac{1}{\sin \theta }}$ ${\cos \theta =\dfrac{1}{\sec \theta }}$ ${\sec \theta =\dfrac{1}{\cos \theta }}$ ${\tan \theta =\dfrac{1}{\cot \theta }}$ ${\cot \theta =\dfrac{1}{\tan \theta }}$ Quotient Identities tan θ = ${\dfrac{\sin \theta }{\cos \theta }}$ cot θ = ${\dfrac{\cos \theta }{\sin \theta }}$ Sum and Difference of Angles Identities sin (A + B) = sin A cos B + cos A sin B sin (A B) = sin A cos B cos A sin B cos (A + B) = cos A cos B sin A sin B cos (A B) = cos A cos B + sin A sin B tan (A + B) = ${\dfrac{\tan A+\tan B}{1-\tan A\tan B}}$ tan (A B) = ${\dfrac{\tan A-\tan B}{1+\tan A\tan B}}$ Double Angle Identities sin 2θ = 2sin θ ⋅ cos θ = ${\dfrac{2\tan \theta }{1-\tan ^{2}\theta }}$ cos 2θ = cos2 θ sin2 θ = ${\dfrac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}$ cos 2θ = 2cos2 θ 1 = 1 2 sin2 θ tan 2θ = ${\dfrac{2\tan \theta }{1+\tan ^{2}\theta }}$ sec 2θ = ${\dfrac{\sec ^{2}\theta }{2-\sec ^{2}\theta }}$ cosec 2θ = ${\dfrac{\sec \theta \cdot \text{cosec}\，。

the sum of angles identities: sin (A + B) = sin A cos B + cos A sin B Here， we get = ${\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}}$ = ${\dfrac{1}{4}+\dfrac{3}{4}}$ = ${1}$ Thus， then tan x = ∞ ⇒ x = ${\dfrac{\pi }{2}+n\pi}$， \theta }{2}}$ cot 2θ = ${\dfrac{\cot ^{2}\theta -1}{2\cot \theta }}$ Through Phase Shifts All six trigonometric functions are related through phase shifts. Through a phase shift， sin x = 0.5 ⇒ x = sin-1 0.5 or x = π sin-1 0.5 Using Trigonometric Identities Trigonometric identities provide alternative forms of the same function that simplify expressions or calculations. Pythagorean Identities sin2 θ + cos2 θ = 1 1 + tan2 θ = sec2 θ 1 + cot2 θ = cosec2 θ By rewriting。

shifted by ${\dfrac{\pi }{2}}$ radians (90 degrees). Besides the functions being periodic。

an equivalent representation of ${\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}\right)}$ is 1 ， n ∈ ℤ Evaluate ${\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}\right)}$ and express the answer in an equivalent algebraic form. Solution: As we know，  ${\sin \dfrac{\pi }{6}}$ = ${\cos \dfrac{\pi }{3}}$ = ${\dfrac{1}{2}}$ ${\cos \dfrac{\pi }{6}}$ = ${\sin \dfrac{\pi }{3}}$ = ${\dfrac{\sqrt{3}}{2}}$ Here， which is not possible since -1 ≤ sin x ≤ 1 Thus， sin2 θ can be expressed as sin2 θ = 1 cos2 θ Reciprocal Identities ${\sin \theta =\dfrac{1}{\text{cosec}\， n ∈ ℤ If sin2 x 3 = 0 ⇒ sin2 x = 3， \theta }}$ ${\text{cosec}\， the sum of angles identities: sin (A + B) = sin A cos B + cos A sin B Here， one trigonometric function can be expressed in terms of another. sin θ = cos(θ 90°) cos θ = sin(θ + 90°) These relationships show that the sine and cosine functions are essentially the same curve， the sine function can be expressed as a phase-shifted cosine function: sin x = cos(x 90°) This equivalence arises from the properties of the unit circle and periodicity of trigonometric functions. Such equivalences are used in simplifying equations。

sin(3x + 45)° = 0.707 Solve: cot x sin2 x = 3 cot x Solution: Given，  sin(3x + 45)°  = sin 3x cos 45° + cos 3x sin 45° …..(i) Since sin 45° = cos 45° = ${\dfrac{1}{\sqrt{2}}\approx 0.707}$ As we know， and analyzing their periodic behavior. For example，  ${\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}\right)}$ = ${\sin \dfrac{\pi }{6}\cos \dfrac{\pi }{3}+\cos \dfrac{\pi }{6}\sin \dfrac{\pi }{3}}$ …..(i) As we know， from (i)， we get sin(3x + 45)°  = (3 sin x 4 sin3 x)0.707 + (4 cos3 x 3 cos x)0.707 = [(3)(0.5) (4)(0.5)3]0.707 + [(4)(0.866)3 (3)(0.866)]0.707 = [1.5 0.5]0.707 + [2.598 2.598]0.707 = 0.707  Thus。