Datrys ar gyfer b
b=\frac{\left(\tan(a)\right)^{2}}{\left(\cos(a)\right)^{2}}
\exists n_{1}\in \mathrm{Z}\text{ : }\left(\left(a\geq 2\pi n_{1}\text{ and }a<2\pi n_{1}+\frac{\pi }{2}\right)\text{ or }\left(a>2\pi n_{1}+\frac{\pi }{2}\text{ and }a\leq 2\pi n_{1}+\pi \right)\right)
Datrys ar gyfer a
\left\{\begin{matrix}a=2\pi n_{2}+\arcsin(\frac{\sqrt{4b+1}-1}{2\sqrt{b}})\text{, }n_{2}\in \mathrm{Z}\text{; }a=2\pi n_{3}-\arcsin(\frac{\sqrt{4b+1}-1}{2\sqrt{b}})+\pi \text{, }n_{3}\in \mathrm{Z}\text{, }&b>0\\a=\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&b=0\end{matrix}\right.
Rhannu
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\sqrt{b}\left(1-\left(\sin(a)\right)^{2}\right)=\sin(a)
Cyfnewidiwch yr ochrau fel bod yr holl dermau newidiol ar yr ochr chwith.
\sqrt{b}-\sqrt{b}\left(\sin(a)\right)^{2}=\sin(a)
Defnyddio’r briodwedd ddosbarthu i luosi \sqrt{b} â 1-\left(\sin(a)\right)^{2}.
\left(1-\left(\sin(a)\right)^{2}\right)\sqrt{b}=\sin(a)
Cyfuno pob term sy'n cynnwys b.
\frac{\left(-\left(\sin(a)\right)^{2}+1\right)\sqrt{b}}{-\left(\sin(a)\right)^{2}+1}=\frac{\sin(a)}{-\left(\sin(a)\right)^{2}+1}
Rhannu’r ddwy ochr â 1-\left(\sin(a)\right)^{2}.
\sqrt{b}=\frac{\sin(a)}{-\left(\sin(a)\right)^{2}+1}
Mae rhannu â 1-\left(\sin(a)\right)^{2} yn dad-wneud lluosi â 1-\left(\sin(a)\right)^{2}.
\sqrt{b}=\frac{\tan(a)}{\cos(a)}
Rhannwch \sin(a) â 1-\left(\sin(a)\right)^{2}.
b=\frac{\left(\tan(a)\right)^{2}}{\left(\cos(a)\right)^{2}}
Sgwariwch ddwy ochr yr hafaliad.
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