tan 2A in Terms of A

We will learn to express trigonometric function of tan 2A in terms of A or tan 2A in terms of tan A. We know if A is a given angle then 2A is known as multiple angles.

How to proof the formula of tan 2A is equals  \(\frac{2  tan  A}{1  -  tan^{2}  A}\)?

We know that for two real numbers or angles A and B,

tan (A + B) =  \(\frac{tan  A  +  tan  B}{1  -  tan  A  tan  B }\)

Now, putting B = A on both sides of the above formula we get,

tan (A + A) =  \(\frac{tan  A  +  tan  A}{1  -  tan  A  tan  A }\)

⇒ tan 2A =  \(\frac{2  tan  A}{1  -  tan^{2}  A}\)

Note: (i) In the above formula we should note that the angle on the R.H.S. is half of the angle on L.H.S. Therefore, tan 60° = \(\frac{2  tan  30°}{1  -  tan^{2}  30°}\).

(ii) The above formula is also known as double angle formulae for tan 2A.


Now, we will apply the formula of multiple angle of tan 2A in terms of A or tan 2A in terms of tan A to solve the below problem.

1. Express tan 4A in terms of tan A


tan 4a

= tan (2 ∙ 2A)

= \(\frac{2  tan  2A}{1 -  tan^{2}  (2A)}\), [Since we know \(\frac{2  tan  A}{1  -  tan^{2}  A}\)]

= \(\frac{2 \cdot \frac{2  tan  A}{1  -  tan^{2}  A}}{1  - (\frac{2  tan  A}{1  -  tan^{2} A})^{2}}\)

= \(\frac{4  tan  A (1  -  tan^{2} A)}{(1  -  tan^{2} A)^{2}  -  4  tan^{2}  A}\)

= \(\frac{4  tan A (1  -  tan^{2}  A)}{1  -  6  tan^{2} A  +  4 tan^{4}}\)

11 and 12 Grade Math

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