Kanuni za pembe maradufu, mara tatu, na nusu

TL;DR

Kanuni za pembe maradufu

• \[\sin 2\alpha = 2\sin \alpha \cos \alpha\]
• \[\begin{align*} \cos 2\alpha &= \cos^{2}\alpha - \sin^{2}\alpha \\ &= 2\cos^{2}\alpha - 1 \\ &= 1 - 2\sin^{2}\alpha \end{align*}\]
• \[\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^{2}\alpha}\]

Kanuni za pembe mara tatu

• \[\sin 3\alpha = 3\sin \alpha - 4\sin^{3}\alpha\]
• \[\cos 3\alpha = 4\cos^{3}\alpha - 3\cos \alpha\]

Kanuni za pembe nusu

• \[\sin^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{2}\]
• \[\cos^{2}\frac{\alpha}{2} = \frac{1 + \cos \alpha}{2}\]
• \[\tan^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{1 + \cos\alpha}\]
• \[\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}\]

Maarifa ya awali

• Kanuni za kujumlisha za trigonometria

Kanuni za pembe za mara nyingi

Kanuni za pembe maradufu

• \[\sin 2\alpha = 2\sin \alpha \cos \alpha\]
• \[\begin{align*} \cos 2\alpha &= \cos^{2}\alpha - \sin^{2}\alpha \\ &= 2\cos^{2}\alpha - 1 \\ &= 1 - 2\sin^{2}\alpha \end{align*}\]
• \[\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^{2}\alpha}\]

Utoaji

Kutoka kwa kanuni za kujumlisha za trigonometria, tunaweza kupata kanuni za pembe maradufu.

\[\begin{gather} \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \label{eqn:sin_add} \\ \cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \label{eqn:cos_add} \\ \tan ( \alpha + \beta ) = \frac { \tan \alpha + \tan \beta } { 1 - \tan \alpha \tan \beta } \label{eqn:tan_add} \end{gather}\]

Tukiweka $\alpha$ badala ya $\beta$,

katika fomula ($\ref{eqn:sin_add}$),

\[\sin 2\alpha = 2\sin \alpha \cos \alpha\]

katika fomula ($\ref{eqn:cos_add}$),

\[\begin{align*} \cos 2 \alpha &= \cos ^ { 2 } \alpha - \sin ^ { 2 } \alpha \\ &= 2 \cos ^ { 2 } \alpha - 1 \\ &= 1 - 2 \sin ^ { 2 } \alpha \end{align*}\]

na katika fomula ($\ref{eqn:tan_add}$),

\[\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^{2} \alpha}\]

Kanuni za pembe mara tatu

• \[\sin 3\alpha = 3\sin \alpha - 4\sin^{3}\alpha\]
• \[\cos 3\alpha = 4\cos^{3}\alpha - 3\cos \alpha\]

Utoaji

Kwa kutumia $\sin 2\alpha = 2\sin\alpha \cos\alpha$ na $\cos 2 \alpha = 1 - 2\sin^{2}\alpha$,

\[\begin{align*} \sin 3 \alpha &= \sin ( \alpha + 2 \alpha ) = \sin \alpha \cos 2 \alpha + \cos \alpha \sin 2 \alpha \\ &= \sin \alpha ( 1 - 2 \sin ^ { 2 } \alpha ) + \cos \alpha ( 2 \sin \alpha \cos \alpha ) \\ &= \sin a ( 1 - 2 \sin ^ { 2 } \alpha ) + 2 \sin \alpha ( 1 - \sin ^ { 2 } \alpha ) \\ &= 3 \sin \alpha - 4 \sin ^ { 3 } \alpha . \end{align*}\]

Kwa njia hiyo hiyo, tukitumia $\sin 2\alpha = 2\sin\alpha \cos\alpha$ na $\cos 2 \alpha = 2\cos^{2}\alpha - 1$,

\[\begin{align*} \cos 3 \alpha &= \cos ( \alpha + 2 \alpha ) = \cos \alpha \cos 2 \alpha - \sin \alpha \sin 2 \alpha \\ &= \cos \alpha ( 2 \cos ^ { 2 } \alpha - 1 ) - \sin \alpha ( 2 \sin \alpha \cos \alpha ) \\ &= \cos \alpha ( 2 \cos ^ { 2 } \alpha - 1 ) - 2 \cos \alpha ( 1 - \cos ^ { 2 } \alpha ) \\ &= 4 \cos ^ { 3 } \alpha - 3 \cos \alpha \end{align*}\]

Kanuni za pembe nusu

• \[\sin^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{2}\]
• \[\cos^{2}\frac{\alpha}{2} = \frac{1 + \cos \alpha}{2}\]
• \[\tan^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{1 + \cos\alpha}\]
• \[\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}\]

Utoaji

Katika kanuni ya pembe maradufu $\cos 2\alpha = 2\cos^{2}\alpha - 1 = 1 - 2\sin^{2}\alpha$, tukiweka $\frac{\alpha}{2}$ badala ya $\alpha$,

\[\cos \alpha = 1 - 2\sin^{2}\frac{\alpha}{2} = 2 \cos^{2}\frac{\alpha}{2} - 1 .\]

Kutoka kwa $ \cos \alpha = 1 - 2\sin^{2}\frac{\alpha}{2} $,

\[\sin^{2}\frac{\alpha}{2}=\frac{1-\cos \alpha}{2} .\]

Kutoka kwa $ \cos \alpha = 2 \cos^{2}\frac{\alpha}{2} - 1 $,

\[\cos^{2}\frac{\alpha}{2}=\frac{1+\cos \alpha}{2} .\]

Kutokana na hili,

\[\tan ^ { 2 } \frac { \alpha } { 2 } = \left . \left( \sin ^ { 2 } \frac{\alpha}{2}\right) \middle/ \left( \cos ^ { 2 } \frac { \alpha } { 2 } \right) \right . = \frac { 1 - \cos \alpha } { 1 + \cos \alpha }\]

inaweza kuonyeshwa, na pia

\[\tan \frac { \alpha } { 2 } = \frac { \sin \frac { \alpha } { 2 } } { \cos \frac { \alpha } { 2 } } = \frac { 2 \sin \frac { \alpha } { 2 } \cos \frac { \alpha } { 2 } } { 2 \cos ^ { 2 } \frac { \alpha } { 2 } } = \frac { \sin \alpha } { 1 + \cos \alpha }\]

pia ni kweli.