Trigonometric Functions Introduction
![trigno_angle](/images/stories/trigno_angle.png)
1. Cos (A + B) = Cos A Cos B – Sin A Sin B
2. Cos (A – B) = Cos A Cos B + Sin A Sin B
3. Sin (A + B) = Sin A Cos B + Cos A Sin B
4. Sin (A – B) = Sin A Cos B - Cos A Sin B
5. Tan (A + B) = Tan A +Tan B1-Tan A Tan B
6. Tan (A – B) = Tan A - Tan B1+Tan A Tan B
Trigonometric Ratios of Angle 2A in terms of Angle A
• Sin 2A = 2 Sin A Cos A
As we know, Sin (A + B) = Sin A Cos B + Cos A Sin B
Replacing B by A in the above equation
Sin (A + A) = Sin A Cos A + Cos A Sin A
Sin 2A = 2 Sin A Cos A
• Cos 2A = Cos2 A – Sin2 A
= 2 Cos2 A – 1
= 1 – 2 Sin2 A
As we know, Cos (A + B) = Cos A Cos B – Sin A Sin B
Replacing B by A in the above equation
Cos (A + A) = Cos A Cos A – Sin A Sin A
Cos 2 A = Cos2 A – Sin2 A
- Cos 2 A = Cos2 A – Sin2 A
Cos 2 A = (1 – Sin2 A – Sin2 A)
Cos 2 A = 1 – 2 Sin2 A
- Cos 2 A = Cos2 A – Sin2 A
Cos 2 A = Cos2 A – (1 – Cos2 A)
Cos 2 A = 2 Cos2 A – 1
• Tan 2A = 2 Tan A
1 – Tan2 A
As we know, Tan (A + B) = Tan A +Tan B1-Tan A Tan B
Replacing B by A in the above equation
Tan (A + A) = Tan A +Tan A1-Tan A Tan A
Tan 2A = 2 Tan A
1 – Tan2 A
Trigonometric Ratios of Angle 3A in terms of Angle A
As we know, Sin (A + B) = Sin A Cos B + Cos A Sin B
Replacing B by 2A in the above equation
Sin (A + 2A) = Sin A Cos 2A + Cos A Sin 2A
Sin 3A = Sin A (1 – 2 Sin2 A) + Cos A (2 Sin A Cos A)
Sin 3A = Sin A – 2 Sin3 A + 2 Sin A Cos2 A
Sin 3A = Sin A – 2 Sin3 A + 2 Sin A (1 – Sin2 A)
Sin 3A = Sin A – 2 Sin3 A + 2 Sin A – 2 Sin2 A
Sin 3A = 3 Sin A – 4 Sin3 A
As we know, Cos (A + B) = Cos A Cos B – Sin A Sin B
Replacing B by 2A in the above equation
Cos (A + 2A) = Cos A Cos 2A – Sin A Sin 2A
Cos 3A = Cos A (2 Cos2 A – 1) – Sin A (2 Sin A Cos A)
Cos 3A = 2 Cos3 A – Cos A – 2 Sin2 A Cos A
Cos 3A = 2 Cos3 A – Cos A – 2 (1 – Cos2 A) Cos A
Cos 3A = 2 Cos3 A – Cos A – 2 Cos A + 2 Cos3 A
Cos 3A = 4 Cos3 A – 3 Cos A
• Tan 3A = 3 Tan A – Tan3 A
1 – 3 Tan2 A
As we know, Tan (A + B) = Tan A +Tan B1-Tan A Tan B
Replacing B by 2A in the above equation
Tan (A + 2A) = Tan A +Tan 2A1-Tan A Tan 2A
• Tan 3A = Tan A + 2 Tan A
1 – Tan2 A
1 – Tan A 2 Tan A
1 – Tan2 A
On Solving we get:
Tan 3A = 3 Tan A – Tan3 A
1 – 3 Tan2 A
Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor!
About eAge Tutoring:
eAgeTutor.com is the premium online tutoring provider. Using materials developed by highly qualified educators and leading content developers, a team of top-notch software experts, and a group of passionate educators, eAgeTutor works to ensure the success and satisfaction of all of its students.
Contact us today to learn more about our tutoring programs and discuss how we can help make the dreams of the student in your life come true!