Uundaji wa Jumla ya Kazi za Trigonometria (Harmonic Addition Theorem)
Kwa jumla ya kazi za trigonometria ya umbo la f(θ) = a cos θ + b sin θ, tunaangalia jinsi ya kupata kazi moja inayolingana nayo, yaani r sin(θ+α) au r cos(θ-β).
Kwa Ufupi
Uundaji wa Jumla ya Kazi za Trigonometria (Harmonic Addition Theorem)
\[(ambapo,\ \cos \alpha = \frac{a}{\sqrt{a^{2}+b^{2}}},\ \sin \alpha = \frac{b}{\sqrt{a^{2}+b^{2}}})\]
- \[a\sin \theta + b\cos \theta = \sqrt{a^{2}+b^{2}} \sin(\theta+\alpha)\]
\[(ambapo,\ \cos \beta = \frac{b}{\sqrt{a^{2}+b^{2}}},\ \sin \beta = \frac{a}{\sqrt{a^{2}+b^{2}}})\]
- \[a\sin \theta + b\cos \theta = \sqrt{a^{2}+b^{2}} \cos(\theta-\beta)\]
Mahitaji ya Awali
Uundaji wa Jumla ya Kazi za Trigonometria (Harmonic Addition Theorem)
Kwa kazi $f(\theta)$ iliyo katika umbo la jumla ya kazi za trigonometria kama $f(\theta) = a \cos \theta + b \sin \theta$, daima zipo nambari halisi $\alpha$, $\beta$ zinazotosheleza $f(\theta)=\sqrt{a^2+b^2} \sin(\theta+\alpha) = \sqrt{a^2+b^2} \cos(\theta-\beta)$.
Kama kwenye mchoro, tukichukua nukta $P(a,b)$ kwenye ndege ya uratibu, na tukisema ukubwa wa pembe inayoundwa na kipande cha mstari $\overline{OP}$ na mwelekeo chanya wa mhimili wa $x$ ni $\alpha$, basi
\[\overline{OP} = \sqrt{a^2+b^2}\]na
\[\cos \alpha = \frac{a}{\sqrt{a^{2} + b^{2}}},\ \sin \alpha = \frac{b}{\sqrt{a^{2} + b^{2}}} \tag{1}\]ni kweli. Wakati huu,
\[\begin{align*} a \sin \theta + b \cos \theta &= \sqrt{a^{2}+b^{2}} \left(\frac{a}{\sqrt{a^{2}+b^{2}}}\sin \theta + \frac{b}{\sqrt{a^{2}+b^{2}}}\cos \theta \right) \\ &= \sqrt{a^{2}+b^{2}}(\cos \alpha \sin \theta + \sin \alpha \cos \theta) \\ &= \sqrt{a^{2}+b^{2}} \sin(\theta + \alpha). \tag{2} \end{align*}\]Kwa njia hiyo hiyo, tukichukua nukta $P^{\prime}(b,a)$ na tukisema ukubwa wa pembe inayoundwa na kipande cha mstari $\overline{OP^{\prime}}$ na mwelekeo chanya wa mhimili wa $x$ ni $\beta$, tunapata yafuatayo.
\[a \sin \theta + b \cos \theta = \sqrt{a^{2}+b^{2}}\cos(\theta-\beta). \tag{3}\] \[Ambapo,\ \cos \beta = \frac{b}{\sqrt{a^{2}+b^{2}}},\ \sin \beta = \frac{a}{\sqrt{a^{2}+b^{2}}}.\]Kwa namna hii, kubadilisha kazi ya trigonometria ya umbo la $a \sin \theta + b \sin \theta$ kuwa katika umbo la $r\sin(\theta+\alpha)$ au $r\cos(\theta-\beta)$ kunaitwa uundaji wa jumla ya kazi za trigonometria (Harmonic Addition).
Mfano
Ikiwa kazi ni $f(\theta)=-\sqrt{3}\sin \theta + \cos \left(\theta - \frac{\pi}{3} \right)$, tafuta thamani kubwa zaidi na thamani ndogo zaidi ya kazi $f(\theta)$ katika sehemu $[0, 2\pi]$.
1. Badilisha iwe katika umbo la $a\sin\theta + b\cos\theta$
Kwa kutumia Kanuni za Kuongeza za Trigonometria, tunaweza kubadilisha fomyula ya kazi tuliyopewa kama ifuatavyo:
\[\begin{align*} f(\theta) &= -\sqrt{3}\sin \theta + \cos \left(\theta - \frac{\pi}{3} \right) \\ &= -\sqrt{3}\sin \theta + \left( \cos\theta \cos\frac{\pi}{3} + \sin\theta \sin\frac{\pi}{3} \right) \\ &= -\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta . \end{align*}\]2. Badilisha iwe katika umbo la $r\sin(\theta+\alpha)$
Tukiweka $a=-\frac{\sqrt{3}}{2}$, $b=\frac{1}{2}$, basi
\[r = \sqrt{a^2+b^2} = \sqrt{\frac{3}{4}+\frac{1}{4}} = 1\]ndivyo ilivyo.
Pia, ipo thamani moja ya nambari halisi $\alpha$ inayotosheleza $0 \leq \alpha<2\pi$, $\cos\alpha = a$, na $\sin\alpha = b$. Kutokana na thamani za uwiano wa trigonometria kwa pembe maalumu, tunaweza kujua kuwa $\alpha = \frac{5}{6}\pi$.
Kwa hiyo, tukibadilisha kazi tuliyopewa $f(\theta)$ kuwa katika umbo la $r\sin(\theta+\alpha)$, tunapata yafuatayo.
\[f(\theta) = \sin \left(\theta + \frac{5\pi}{6} \right).\]3. Tafuta thamani kubwa zaidi na ndogo zaidi katika sehemu iliyotolewa
Kazi $f(\theta) = \sin \left(\theta + \frac{5\pi}{6} \right)$ ni kazi ya kipindi yenye kipindi cha $2\pi$, na katika sehemu iliyotolewa ina thamani kubwa zaidi $1$ na thamani ndogo zaidi $-1$.
\[\therefore M=1,\ m=-1\]