Question

Deduce the identity of :
1−2sin2A

A sin 2A
B tan 2A
C cos 2A
D cos A

Answer 1

Answer on Question #51811 – Math – Trigonometry

Deduce the identity of :

$1 - 2\sin^2 A$

A $\sin 2A$

B $\tan 2A$

C $\cos 2A$

D $\cos A$

Solution

Method 1

This method uses ideas of Trigonometry.

Recall formulas

$$\cos^2 A + \sin^2 A = 1$$

and $\cos(2A) = \cos^2 A - \sin^2 A$, which follows from

$$\cos(A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B.$$

Then

$$\begin{array}{l} 1 - 2\sin^2 A = (\cos^2 A + \sin^2 A) - 2\sin^2 A = \\ = \cos^2 A + (\sin^2 A - 2\sin^2 A) = \cos^2 A - \sin^2 A = \cos(2A) \end{array}$$

Method 2

This method uses ideas of Complex Analysis.

According to Euler's formula $\sin \alpha = \left(e^{i\alpha} - e^{-i\alpha}\right) / 2$,

$$\begin{array}{l} 1 - 2\sin^2 \alpha = 1 - 2 \cdot \left(\frac{\exp[i\alpha] - \exp[-i\alpha]}{2i}\right)^2 = 1 - 2 \frac{\left(\exp[2i\alpha] - 2\exp[i\alpha]\exp[-i\alpha] + \exp[-2i\alpha]\right)}{4i^2} = \\ = 1 + \frac{1}{2} \left(\exp[2i\alpha] - 2 + \exp[-2i\alpha]\right) = \frac{\left(\exp[2i\alpha] + \exp[-2i\alpha]\right)}{2} = \cos 2\alpha \end{array}$$

Answer: C $\cos (2A)$

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