\(\sin{(A\pm B)}=\sin{A}\cos{B}\pm \cos{A}\sin{B}\)
\(\cos{(A\pm B)}=\cos{A}\cos{B}\mp \sin{A}\sin{B}\)
\(\tan{(A\pm B)}=\dfrac{\tan{A}\pm \tan{B}}{1\mp \tan{A}\tan{B}}\)
\(\sin{2A}=2\sin{A}\cos{A}\)
\(\begin{aligned} \cos{2A}&= \cos^2{A}-\sin^2{A} \\ &=2\cos^2{A}-1 \\ &=1-2\sin^2{A} \end{aligned}\)
\(\tan{2A}=\dfrac{2\tan{A}}{1-\tan^2{A}}\)
\(\sin{\dfrac{A}{2}}=\pm\sqrt{\dfrac{1-\cos{A}}{2}}\)
\(\cos{\dfrac{A}{2}}=\pm\sqrt{\dfrac{1+\cos{A}}{2}}\)
\(\tan{\dfrac{A}{2}}=\pm\sqrt{\dfrac{\sin{A}}{1+\cos{A}}}\)
(a)
\(\begin{aligned} \text{sin }(90^{\circ}+A) &= \text{sin }90^{\circ} \text{ cos }A + \text{ cos }90^{\circ} \text{ sin }A\\ &=(1) \text{ cos }A + (0) \text{ sin }A\\ &=\text{ cos }A. \end{aligned}\)
(b)
Recall that:
\(\begin{aligned} \text{tan }15^{\circ}&= \tan{(60^\circ-45^\circ)} \\\\ \text{tan }(60^{\circ} -45^{\circ})&=\dfrac{\text{tan }60^{\circ} - \text{tan }45^{\circ}}{1+\text{tan }60^{\circ} \ \text{tan }45^{\circ}}\\\\ &=\dfrac{\sqrt 3 - 1}{1+ (\sqrt3)(1)}\\\\ &=\dfrac{\sqrt3 -1}{\sqrt 3+1}\\\\ &=2-\sqrt3. \end{aligned}\)
\(\begin{aligned} 2\sin{15^\circ}\cos{15^\circ}&=\sin{2(15^\circ)} \\ &=\sin{30^\circ} \\ &=\dfrac{1}{2}. \end{aligned}\)
From right-hand side:
\(\begin{aligned} \dfrac{1-\cos{x}}{\sin{x}}&=\dfrac{1-\left( 1-2\sin^2{\dfrac{x}{2}} \right)}{2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}} \\\\ &=\dfrac{2\sin^2{\dfrac{x}{2}}}{2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}} \\\\ &=\dfrac{\sin{\dfrac{x}{2}}}{\cos{\dfrac{x}{2}}} \\\\ &=\tan{\dfrac{x}{2}}. \end{aligned}\)
Hence, it is proven that \(\text{tan }\dfrac{x}{2} = \dfrac{1-\text{ cos }x}{\text{sin }x}\).
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