3) Trigonometrical ratios:
In a right-angled triangle, the Pythagoras theorem states
(perpendicular )2 + ( base )2 = ( hypotenuse )2
There are some properties pertaining to the right-angled triangle. It is to be noted that in the following formulas, P stands for perpendicular, B stands for base and H stands for the hypotenuse.
- SinA = P / H
- CosA = B / H
- TanA = P / B
- CotA = B / P
- CosecA = H / P
- SecA = H/B
- Sin2A + Cos2A = 1
- Tan2A + 1 = Sec2A
Cot2A + 1 = Cosec2A
In order to find a relationship between various trigonometric identities, there are some important formulas:
1. TanA = SinA / CosA
2. CotA = CosA / SinA
3. CosecA = 1 / SinA
4. SecA = 1 / CosA
1. TanA = SinA / CosA
2. CotA = CosA / SinA
3. CosecA = 1 / SinA
4. SecA = 1 / CosA
There are some formulas that are very crucial to solving higher level sums.
- Sin (A +B) =SinA . CosB + CosA . SinB
- Sin (A – B) = SinA . CosB – CosA . SinB
- Cos (A + B) = CosA . CosB – SinA . SinB
- Cos (A – B) = CosA. CosB + SinA . SinB
- Tan (A + B) = TanA + TanB / 1 – TanA . TanB
- Tan (A – B) = TanA –TanB / 1 + TanA . TanB
- Sin ( A + B) . Sin (A – B) = Sin2A – Sin2B = Cos2B – Cos2A
- Cos (A + B) . Cos (A – B) = Cos2A – Sin2B = Cos2B – Sin2A
- Sin2A = 2 . SinA . CosA = 2 . TanA / (1 + Tan2A)
- Cos2A = Cos2A – Sin2A = 1 – 2Sin2A = 2Cos2A – 1 = (1 – Tan2A) / (1 + Tan2A)
- Tan2A = 2TanA / (1 – Tan2A)
- Sin3A = 3 . SinA – 4 . Sin3A
- Cos3A = 4 . Cos3A – 3 . CosA
- Tan3A = (3TanA – Tan3A) / (1 – 3Tan2A)
- SinA + SinB = 2 Sin (A + B)/2 Cos (A – B)/2
- SinA – SinB = 2 Sin (A – B)/2 Cos (A + B)/2
- CosA + CosB = 2 Cos(A – B)/2 Cos (A + B)/2
- CosA – CosB = 2 Sin(B – A)/2 Sin (A + B)/2
- TanA + TanB = Sin (A + B) / CosA . CosB
- SinA CosB = Sin (A + B) + Sin (A – B)
- CosA SinB = Sin (A + B) – Sin (A – B)
- CosA CosB = Cos (A + B) + Cos (A – B)
- SinA SinB = Cos (A – B) – Cos (A + B)
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