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
⇒ sin 2A = 2 \(\frac{sin A}{cos A}\) ∙ cos\(^{2}\) A
⇒ sin 2A = 2 tan A ∙ \(\frac{1}{sec^{2} A}\)
⇒ 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 ≤ A ≤ π / 4)
Solution:
Given, sin 2A = 4/5
Therefore, \(\frac{2 tan A}{1 + tan^{2} A}\) = 4/5
⇒ 4 + 4 tan\(^{2}\) A = 10 tan A
⇒ 4 tan\(^{2}\) A  10 tan A + 4 = 0
⇒ 2 tan\(^{2}\) A  5 tan A + 2 = 0
⇒ 2 tan\(^{2}\) A  4 tan A  tan A + 2 = 0
⇒ 2 tan A (tan A  2)  1 (tan A  2) =0
⇒ (tan A  2) (2 tan A  1) = 0
Therefore, tan A  2 = 0 and 2 tan A  1 = 0
⇒ tan A = 2 and tan A = 1/2
According to the problem, 0 ≤ A ≤ π/4
Therefore, tan A = 2 is impossible
Therefore, the required value of tan A is 1/2.
`11 and 12 Grade Math
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