Introduction to trigonometry
Download as PPTX, PDF0 likes115 views
The document summarizes trigonometric ratios and identities. It provides: 1) Tables of trigonometric ratios for specific angles from 0 to 90 degrees, including sin, cos, tan, cot, sec, and cosec. 2) Examples of solving trigonometric equations by using identities and properties of trig functions. 3) Proofs of several trigonometric identities by manipulating ratios and using Pythagorean and other identities, such as expressing all ratios in terms of one function, like cot.
![5(v) Cos A – Sin A + 1 = Cosec A + Cot A
Cos A + Sin A – 1
L.H.S = Cos A – Sin A + 1
Cos A + Sin A – 1
dividing the Nr & Dr by Sin A, we get
= Cos A/Sin A – Sin A/Sin A + 1/Sin A
Cos A/Sin A + Sin A/Sin A – 1/Sin A
= CotA – 1 + Cosec A
CotA + 1 – Cosec A
= (CotA +Cosec A) - (Cosec A+CotA )( Cosec A-CotA)
CotA + 1 – Cosec A
= (CotA +Cosec A)[ 1 - Cosec A + CotA]
[ 1 - Cosec A + CotA]
= Cosec A + Cot A = R.H.S
Hence it is proved.](https://image.slidesharecdn.com/std10trigonometry-200717173829/85/Introduction-to-trigonometry-14-320.jpg)
![VII) Sinθ - 2Sin3 θ = Tanθ
2C0s3 θ - Cos θ
LHS = Sinθ - 2Sin3 θ
2C0s3 θ - Cos θ
= Sinθ - 2Sin3 θ
2C0s3 θ - Cos θ
= Sinθ( 1 - 2Sin2 θ)
Cos θ (2C0s2 θ - 1 )
= Sinθ [ 1 – 2( 1- Cos2θ)]
Cos θ (2C0s2 θ - 1 )
= Sinθ( 1 – 2 + 2Cos2θ)
Cos θ (2C0s2 θ - 1 )
= Sinθ ( 2Cos2θ - 1 )
Cos θ (2C0s2 θ - 1 )
= Sinθ
Cosθ
= Tanθ
= RHS
Hence it is proved.](https://image.slidesharecdn.com/std10trigonometry-200717173829/85/Introduction-to-trigonometry-16-320.jpg)
![x)(1 + Tan2A)/(1+ Cot2 A)= [(1 - TanA)/(1-CotA)]2 = Tan2A
Solution:
(1 + Tan2A) / (1+ Cot2 A) = Sec2A / Cosec2A
= (1/Cos2A) / ( 1/Sin2A)
= 1/Cos2A x Sin2A/1
= Sin2A /Cos2A = Tan2A
[(1 + TanA) / (1+CotA)]2 = [(1 - SinA/CosA) / (1- CosA/SinA)]2
= [(CosA- SinA)/CosA /(SinA -
CosA)/SinA)]2
= [(CosA+ SinA)/CosA X - (SinA
/(CosA +SinA)]
= ( - SinA / Cos A ) 2 = (-
TanA ) 2 = Tan 2 A therefore,
(1 + Tan2A)/(1+ Cot2 A)= [(1 - TanA)/(1-CotA)]2 = Tan2A
Hence it is proved.](https://image.slidesharecdn.com/std10trigonometry-200717173829/85/Introduction-to-trigonometry-19-320.jpg)
![Prove that:
(1 + Cot2A) / (1+ Tan2 A)
= [(1 - Cot A) / (1- TanA)]2
= Cot2A](https://image.slidesharecdn.com/std10trigonometry-200717173829/85/Introduction-to-trigonometry-20-320.jpg)
More Related Content
Similar to Introduction to trigonometry (20)
![[Medicalstudyzone.com] 1. AIIMS NOV EMBER 2015 SOLVED PAPER.pdf](https://cdn.slidesharecdn.com/ss_thumbnails/medicalstudyzone-250904212000-ef4cbf83-thumbnail.jpg?width=560&fit=bounds)
Introduction to trigonometry
- 1. TRIGONOMETRY Ratios of specific angles 0/30/45/60/90
- 3. RATIOS FOR 30 AND 60
- 4. TRIGONOMETRIC RATIOS OF SPECIFIC ANGLES Angle A 00 300 450 600 900 Sin 0 ½ 1/√2 √3/2 1 Cos 1 √3/2 1/√2 1/2 0 Tan 0 1/√3 1 √3 undefined Cot undefined √3 1 1/√3 0 Sec 1 2/√3 √2 2 undefined Cosec undefined 2 1/√2 2/√3 1
- 6. Qn: If tan (A + B) = √3 and tan (A – B) =1/√3; 0° < A + B < 90°; A > B, find A and B. Ans: Given, tan (A + B) = √3 => tan(A+B) = tan60° => A + B = 60° => 1 tan (A- B) =1/√3 => tan(A- B) = tan30° => A - B = 30° => ② 1 + ② A + B = 60° A - B = 30° 2A = 90° => A = 90°/2 = 45° from ② 45° + B = 60° => B = 60° - 45° = 15° therefore, <A = 45° and <B = 15°
- 7. HOME WORK Qn: 1. If Sin(A + B) = 1 & Cos (A – B) =√3/2; 0° < A + B < 90°; A > B, find A and B. Qn: 2. If Cot(A + B) = 1/√3 & Cosec(A – B) = 2; 0° < A + B < 90°; A > B, find A and B.
- 9. C In ∆ABC, ∟B = 90° By Pythagorous theorem B A AC2 = AB2 + BC2 => AB2 + BC2 = AC2 AC2 AC2 AC2 Sin2A + Cos2A = 1 AC2 = AB2 + BC2 AB2 AB2 AB2 Sec2A = 1 + Tan2A AC2 = AB2 + BC2 BC2 BC2 BC2 Cosec2A = Cot2A + 1
- 10. IDENTITY - 1 Sin2A + Cos2A = 1 => Sin2A = 1 - Cos2A • => SinA = √ 1 - Cos2A => Cos2A = 1 - Sin2A • => Cos A = √ 1 - Sin2A
- 11. IDENTITY - 2 1 + Tan2A = Sec2A => Sec A = √ 1 + Tan2A • => TanA = √ Sec2A - 1 Sec2A - Tan2A = 1 (SecA - TanA) (SecA + TanA) = 1 (SecA - TanA) = 1/ (SecA + TanA) (SecA + TanA) = 1/ (SecA - TanA)
- 12. IDENTITY - 3 1 + Cot2A = Cosec2A => Cosec A = √ 1 + Cot2A • => CotA = √Cosec2A - 1 Cosec2A - Cot2A = 1 (CosecA - CotA) (CosecA + CotA) = 1 (CosecA - CotA) = 1/ (CosecA + CotA) (CosecA + CotA) = 1/ (CosecA - CotA)
- 13. QN 1. EXPRESS ALL THE TRIGONOMETRIC RATIOS IN TERMS OF COTA Cot A Tan A = 1/Cot A CosecA = √ 1 – Cot2A Sin A = 1/ Cosec A = 1/√ 1 – Cot2A Cos A = √ 1 – Sin2A = √ 1 –( 1/√ 1 – Cot2A)2 = √1 –( 1/ 1 – Cot2A)=√(1– Cot2A –1)/ 1 – Cot2A) = √ Cot2A/ (Cot2A-1) = CotA/√(Cot2A-1) Sec A = 1/CosA = 1/( CotA/√(Cot2A-1)) = √(Cot2A-1)/CotA
- 14. 5(v) Cos A – Sin A + 1 = Cosec A + Cot A Cos A + Sin A – 1 L.H.S = Cos A – Sin A + 1 Cos A + Sin A – 1 dividing the Nr & Dr by Sin A, we get = Cos A/Sin A – Sin A/Sin A + 1/Sin A Cos A/Sin A + Sin A/Sin A – 1/Sin A = CotA – 1 + Cosec A CotA + 1 – Cosec A = (CotA +Cosec A) - (Cosec A+CotA )( Cosec A-CotA) CotA + 1 – Cosec A = (CotA +Cosec A)[ 1 - Cosec A + CotA] [ 1 - Cosec A + CotA] = Cosec A + Cot A = R.H.S Hence it is proved.
- 15. Prove that : SinA – Cos A + 1 = Sec A + Tan A Sin A + Cos A – 1
- 16. VII) Sinθ - 2Sin3 θ = Tanθ 2C0s3 θ - Cos θ LHS = Sinθ - 2Sin3 θ 2C0s3 θ - Cos θ = Sinθ - 2Sin3 θ 2C0s3 θ - Cos θ = Sinθ( 1 - 2Sin2 θ) Cos θ (2C0s2 θ - 1 ) = Sinθ [ 1 – 2( 1- Cos2θ)] Cos θ (2C0s2 θ - 1 ) = Sinθ( 1 – 2 + 2Cos2θ) Cos θ (2C0s2 θ - 1 ) = Sinθ ( 2Cos2θ - 1 ) Cos θ (2C0s2 θ - 1 ) = Sinθ Cosθ = Tanθ = RHS Hence it is proved.
- 17. Prove that : Cosθ - 2Cos3 θ = Cotθ 2Sin3 θ - Sin θ
- 18. VIII) (SinA+CosecA) 2 + (CosA + Sec A) 2 = 7 + Tan2A +Cot2A Solution: (SinA+CosecA) 2 = Sin 2 A+Cosec2 A + 2 . SinA.CosecA = Sin 2A+ 1+ Cot2 A + 2 . SinA. 1/Sin A = Sin 2A+ Cot2 A + 3 ------1 (CosA + Sec A) 2 = Cos2 A + Sec2 A + 2 . CosA.SecA = Cos2 A +1 + Tan2A + 2.CosA.1/Cos A = Cos2 A + Tan2A + 3 ------ 2 now, LHS =1+2 = Sin 2A+ Cot2 A+3 + Cos2 A+Tan2A + 3 = 1 + 3 + 3 + Tan2A +Cot2A = 7 + Tan2A +Cot2A = RHS Hence it is proved.
- 19. x)(1 + Tan2A)/(1+ Cot2 A)= [(1 - TanA)/(1-CotA)]2 = Tan2A Solution: (1 + Tan2A) / (1+ Cot2 A) = Sec2A / Cosec2A = (1/Cos2A) / ( 1/Sin2A) = 1/Cos2A x Sin2A/1 = Sin2A /Cos2A = Tan2A [(1 + TanA) / (1+CotA)]2 = [(1 - SinA/CosA) / (1- CosA/SinA)]2 = [(CosA- SinA)/CosA /(SinA - CosA)/SinA)]2 = [(CosA+ SinA)/CosA X - (SinA /(CosA +SinA)] = ( - SinA / Cos A ) 2 = (- TanA ) 2 = Tan 2 A therefore, (1 + Tan2A)/(1+ Cot2 A)= [(1 - TanA)/(1-CotA)]2 = Tan2A Hence it is proved.
- 20. Prove that: (1 + Cot2A) / (1+ Tan2 A) = [(1 - Cot A) / (1- TanA)]2 = Cot2A
- 21. Prove that √( 1 + Sin A ) / √( 1 – SinA ) = Sec A + TanA LHS = √( 1 + Sin A ) / √( 1 – SinA ) Rationalising the denominator = √( 1 + Sin A ) / √( 1 – SinA ) x √( 1 + Sin A ) / √( 1 + SinA ) = √( 1 + Sin A ) √( 1 + Sin A ) / √( 1 – SinA ) √( 1 + Sin A ) = √( 1 + Sin A ) 2 / √( 1 - Sin2 A ) = (1 + Sin A ) / √ Cos2 A = (1 + Sin A ) / Cos A = 1 / CosA + Sin A/ Cos A = Sec A + Tan A = RHS Hence it proved.
- 22. Prove that: √( 1 + Cos A ) / √( 1 – CosA ) = Cosec A + CotA