6.5 Formulae of sin (A±B), cos (A±B), tan (A±B), sin 2A, cos 2A, tan 2A

6.5 Formulae of sin (A ± B), cos (A ± B), tan (A ± B), sin 2A, cos 2A, tan 2A

(A) Compound Angles Formulae:

\begin{matrix} ∙ sin (A \pm B) = sin A cos B \pm cos A sin B ∙ cos (A \pm B) = cos A cos B \mp sin A sin B ∙ tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B} \end{matrix}

$\begin{array}{l} • \sin (A \pm B) = \sin A \cos B \pm \cos A \sin B \\ • \cos (A \pm B) = \cos A \cos B \mp \sin A \sin B \\ • \tan (A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \end{array}$

(B) Double Angle Formulae:

sin 2A = 2 sin A cos A
cos 2A = cos² A – sin² A
cos 2A = 2 cos² A – 1
cos 2A = 1 – 2 sin² A
$tan 2 A = \frac{2 tan A}{1 - {tan}^{2} A}$ $\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}$

(C) Half Angle Formulae:

\begin{matrix} ∙ sin A = 2 sin \frac{A}{2} cos \frac{A}{2} ∙ cos A = {sin}^{2} \frac{A}{2} - {cos}^{2} \frac{A}{2} cos A = 2 {cos}^{2} \frac{A}{2} - 1 cos A = 1 - 2 {cos}^{2} \frac{A}{2} ∙ tan A = \frac{2 tan \frac{A}{2}}{1 - {tan}^{2} \frac{A}{2}} \end{matrix}

$\begin{array}{l} • \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2} \\ • \cos A = \sin^{2} \frac{A}{2} - \cos^{2} \frac{A}{2} \\ \cos A = 2 \cos^{2} \frac{A}{2} - 1 \\ \cos A = 1 - 2 \cos^{2} \frac{A}{2} \\ • \tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^{2} \frac{A}{2}} \end{array}$

5.5.1 Proving Trigonometric Identities using Addition Formula and Double Angle Formulae

Example 1:

Prove each of the following trigonometric identities.

\begin{matrix} (a) \frac{s i n (A + B) - s i n (A - B)}{c o s A cos B} = 2 tan B (b) \frac{cos (A + B)}{sin A cos B} = cot A - tan B (c) tan (A + 45^{o}) = \frac{sin A + cos A}{c o s A - sin A} \end{matrix}

$\begin{array}{l} (a) \frac{s i n (A + B) - s i n (A - B)}{c o s A \cos B} = 2 \tan B \\ (b) \frac{\cos (A + B)}{\sin A \cos B} = \cot A - \tan B \\ (c) \tan (A + 45^{o}) = \frac{\sin A + \cos A}{c o s A - \sin A} \end{array}$

Solution:

(a)

\begin{matrix} L H S = \frac{s i n (A + B) - s i n (A - B)}{c o s A cos B} = \frac{(sin A cos B + cos A sin B) - (sin A cos B - cos A sin B)}{c o s A cos B} = \frac{2 cos A sin B}{cos A cos B} = \frac{2 sin B}{cos B} = 2 tan B = R H S (proven) \end{matrix}

$\begin{array}{l} L H S \\ = \frac{s i n (A + B) - s i n (A - B)}{c o s A \cos B} \\ = \frac{(\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B)}{c o s A \cos B} \\ = \frac{2 \cos A \sin B}{\cos A \cos B} \\ = \frac{2 \sin B}{\cos B} \\ = 2 \tan B = R H S (proven) \end{array}$

(b)

\begin{matrix} L H S = \frac{cos (A + B)}{sin A cos B} = \frac{cos A cos B - sin A sin B}{sin A cos B} = \frac{cos A cos B}{sin A cos B} - \frac{sin A sin B}{sin A cos B} = \frac{cos A}{sin A} - \frac{sin B}{cos B} = cot A - tan B = R H S (proven) \end{matrix}

$\begin{array}{l} L H S \\ = \frac{\cos (A + B)}{\sin A \cos B} \\ = \frac{\cos A \cos B - \sin A \sin B}{\sin A \cos B} \\ = \frac{\cos A \cos B}{\sin A \cos B} - \frac{\sin A \sin B}{\sin A \cos B} \\ = \frac{\cos A}{\sin A} - \frac{\sin B}{\cos B} \\ = \cot A - \tan B \\ = R H S (proven) \end{array}$

(c)

\begin{matrix} L H S = tan (A + 45^{o}) = \frac{t a n A + tan 45^{o}}{1 - t a n A tan 45^{o}} = \frac{t a n A + 1}{1 - t a n A} \leftarrow tan 45^{o} = 1 = \frac{\frac{sin A}{cos A} + 1}{1 - \frac{sin A}{cos A}} = \frac{sin A + cos A}{cos A} \times \frac{cos A}{cos A - sin A} = \frac{sin A + cos A}{cos A - sin A} = R H S (proven) \end{matrix}

$\begin{array}{l} L H S \\ = \tan (A + 45^{o}) \\ = \frac{t a n A + \tan 45^{o}}{1 - t a n A \tan 45^{o}} \\ = \frac{t a n A + 1}{1 - t a n A} \leftarrow \tan 45^{o} = 1 \\ = \frac{\frac{\sin A}{\cos A} + 1}{1 - \frac{\sin A}{\cos A}} \\ = \frac{\sin A + \cos A}{\cos A} \times \frac{\cos A}{\cos A - \sin A} \\ = \frac{\sin A + \cos A}{\cos A - \sin A} \\ = R H S (proven) \end{array}$

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