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sin 2A in Terms of tan A
We will learn how to express the multiple angle of sin 2A in terms of tan A.
Trigonometric function of sin 2A in terms of tan A is also known as one of the double angle formula.
We know if A is a number or angle then we have,
sin 2A = 2 sin A cos A
β sin 2A = 2 \(\frac{sin A}{cos A}\) β cos\(^{2}\) A
β sin 2A = 2 tan A β \(\frac{1}{sec^{2} A}\)
β sin 2A = \(\frac{2 tan A}{1 + tan^{2} A}\)
There for sin 2A = \(\frac{2 tan A}{1 + tan^{2} A}\)
Now, we will apply the
formula of multiple angle of sin 2A in terms of tan A to solve the below problem.
1. If sin 2A = 4/5 find the value of tan A (0 β€ A β€ Ο / 4)
Solution:
Given, sin 2A = 4/5
Therefore, \(\frac{2 tan A}{1 + tan^{2} A}\) = 4/5
β 4 + 4 tan\(^{2}\) A = 10 tan A
β 4 tan\(^{2}\) A - 10 tan A + 4 = 0
β 2 tan\(^{2}\) A - 5 tan A + 2 = 0
β 2 tan\(^{2}\) A - 4 tan A - tan A + 2 = 0
β 2 tan A (tan A - 2) - 1 (tan A - 2) =0
β (tan A - 2) (2 tan A - 1) = 0
Therefore, tan A - 2 = 0 and 2 tan A - 1 = 0
β tan A = 2 and tan A = 1/2
According to the problem, 0 β€ A β€ Ο/4
Therefore, tan A = 2 is impossible
Therefore, the required value of tan A is 1/2.
β Multiple Angles
- sin 2A in Terms of A
- cos 2A in Terms of A
- tan 2A in Terms of A
- sin 2A in Terms of tan A
- cos 2A in Terms of tan A
- Trigonometric Functions of A in Terms of cos 2A
- sin 3A in Terms of A
- cos 3A in Terms of A
- tan 3A in Terms of A
- Multiple Angle Formulae
11 and 12 Grade Math
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