Trigonometry is an important part of mathematics. It involves its own functions sine, cosine, etc. They are essential tools for simplifying expressions and solving trigonometric equations.
In this article, we will learn about trigonometric identities, their important formulas, and how to solve related equations.
What are Trigonometric Identities?
Trigonometric identities are mathematical equations involving trigonometric functions (like sine, cosine, and tangent) that hold for all values of the variables within their domains. They are essential tools for simplifying expressions and solving trigonometric equations.
Important Formulas
Various important trigonometric identities are:
- sin2 θ + cos2 θ = 1
- tan2 θ = sec2 θ – 1
- cosec2θ = 1+ cot2θ
- sin (-θ) = – Sin θ
- cos (-θ) = Cos θ
- tan (-θ) = – Tan θ
- cot (-θ) = – Cot θ
- sec (-θ) = Sec θ
- cosec (-θ) = -Cosec θ
- sin(x+y) = sin(x)cos(y)+cos(x)sin(y)
- cos(x+y) = cos(x)cos(y)–sin(x)sin(y)
- sin(x–y) = sin(x)cos(y)–cos(x)sin(y)
- cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
- cos(2x) = 2cos2(x)-1 = 1–2sin2(x)= cos2(x)-sin2(x)
- Sin 3x = 3sin x – 4sin3x
- Cos 3x = 4cos3x-3cos x
Inverse Trigonometry Formulas
- sin-1 (–x) = – sin-1 x
- cos-1 (–x) = π – cos-1 x
- tan-1 (–x) = – tan-1 x
- cosec-1 (–x) = – cosec-1 x
- sec-1 (–x) = π – sec-1 x
- cot-1 (–x) = π – cot-1 x
Trigonometric Identities Practice Questions
Question 1: Prove that (sin4θ – cos4θ +1) cosec2θ = 2
Solution:
L.H.S. = (sin4θ – cos4θ +1) cosec2θ
= [(sin2θ – cos2θ) (sin2θ + cos2θ) + 1] cosec2θ
Using the identity sin2A + cos2A = 1,
= (sin2θ – cos2θ + 1) cosec2θ
= [sin2θ – (1 – sin2θ) + 1] cosec2θ
= 2 sin2θ cosec2θ
= 2 sin2θ (1/sin2θ)
= 2
= RHS
Question 2: Solve expression: cosec2θ - cot2θ.
Solution:
1 + cot2θ = cosec2θ
cosec2θ-cot2θ = 1+cot2θ-cot2θ
= 1 = R.H.S.
Question 3: (1 - sin A)/(1 + sin A) = (sec A - tan A)2
Solution:
L.H.S = (1 - sin A)/(1 + sin A)
= (1 - sin A)2/(1 - sin A) (1 + sin A),[Multiply both numerator and denominator by (1 - sin A)
= (1 - sin A)2/(1 - sin2 A)
= (1 - sin A)2/(cos2 A), [Since sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 - sin2 θ]
= {(1 - sin A)/cos A}2
= (1/cos A - sin A/cos A)2
= (sec A – tan A)2 = R.H.S.
Question 4: tan4 θ + tan2 θ = sec4 θ - sec2 θ
Solution:
L.H.S = tan4 θ + tan2 θ
= tan2 θ (tan2 θ + 1)
= (sec2 θ - 1) (tan2 θ + 1) [since, tan2 θ = sec2 θ – 1]
= (sec2 θ - 1) sec2 θ [since, tan2 θ + 1 = sec2 θ]
= sec4 θ - sec2 θ = R.H.S.
Question 5: (tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ
Solution:
L.H.S = (tan θ + sec θ - 1)/(tan θ - sec θ + 1)
= [(tan θ + sec θ) - (sec2 θ - tan2 θ)]/(tan θ - sec θ + 1), [Since, sec2 θ - tan2 θ = 1]
= {(tan θ + sec θ) - (sec θ + tan θ) (sec θ - tan θ)}/(tan θ - sec θ + 1)
= {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1)
= {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1)
= tan θ + sec θ
= (sin θ/cos θ) + (1/cos θ)
= (sin θ + 1)/cos θ
= (1 + sin θ)/cos θ = R.H.S
Question 6: (1 + cos θ)/sin θ + sin θ/(1 + cos θ) = 2 cosec θ
Solution:
L.H.S = (1 + cos θ)/sin θ + sin θ/(1 + cos θ)
= (1 + cos θ)2/[sin θ(1 + cos θ)] + sin θ2/[sin θ(1 + cos θ)]
= [(1 + cos θ)2 + sin θ2]/[sin θ(1 + cos θ)]
= [1 + 2 cos θ + cos2 θ + sin2 θ]/[sin θ(1 + cos θ)]
= [1 + 2 cos θ + 1]/[sin θ(1 + cos θ)]; [since cos2 θ + sin2 θ = 1]
= [2 + 2 cos θ]/[sin θ(1 + cos θ)]
= 2(1 + cos θ)/[sin θ(1 + cos θ)]
= 2/sin θ
= 2 cosec θ = R.H.S. (Proved)
Question 7: Prove cosθ/(1+sinθ) = (1-sinθ)/cosθ.
Solution:
cosθ/(1+sinθ){(1-sinθ)/(1-sinθ)} {multiplying and dividing by 1-sinθ on LHS}
= cosθ(1-sinθ)/1-sin2θ
= cosθ(1-sinθ)/cos2θ
= 1-sinθ/cosθ = R.H.S.
Worksheet: Trigonometric Identities
Problem 1: Prove it sin2 θ + cos2 θ = 1.
Problem 2: Prove it tan2 θ = sec2 θ – 1.
Problem 3: Prove it sin(x+y) = sin(x)cos(y)+cos(x)sin(y).
Problem 4: Prove it cos(x+y) = cos(x)cos(y)–sin(x)sin(y).
Problem 5: Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using the identity cosec2A = 1 + cot2A.
Problem 6: Prove it sin(x–y) = sin(x)cos(y)–cos(x)sin(y).
Problem 7: Prove it cos(x–y) = cos(x)cos(y) + sin(x)sin(y).
Problem 8: If a cos3α + 3a cos α sin2α = m and a sin3α + 3a cos2α sin α = n, then find (m + n)2/3 + (m – n)2/3.
Problem 9: Prove it cos(2x) = 2cos2(x)-1 = 1–2sin2(x).
Problem 10: Prove it Sin 3x = 3sin x – 4sin3x.
Problem 11: Prove it Cos 3x = 4cos3x-3cos x.
Problem 12: Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A).
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