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.

**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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