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
Solution:
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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