Whakaoti i te cos(2*30^{circ})=1-tan^230^circ/1+tan^230^circ

$$\cos ( 2 \times 30 ^ { \circ } ) = \frac { 1 - \tan ^ { 2 } 30 ^ { \circ } } { 1 + \tan ^ { 2 } 30 ^ { \circ } }$$

Manatoko

Ngā Upane Otinga

Whakareatia te 2 ki te 30, ka 60.

$$\cos(60)=\frac{1-\left(\tan(30)\right)^{2}}{1+\left(\tan(30)\right)^{2}}$$

Get the value of \cos(60) from trigonometric values table.

$$\frac{1}{2}=\frac{1-\left(\tan(30)\right)^{2}}{1+\left(\tan(30)\right)^{2}}$$

Get the value of \tan(30) from trigonometric values table.

$$\frac{1}{2}=\frac{1-\left(\frac{\sqrt{3}}{3}\right)^{2}}{1+\left(\tan(30)\right)^{2}}$$

Kia whakarewa i te \frac{\sqrt{3}}{3} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$\frac{1}{2}=\frac{1-\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}{1+\left(\tan(30)\right)^{2}}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{1}{2}=\frac{1-\frac{3}{3^{2}}}{1+\left(\tan(30)\right)^{2}}$$

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$\frac{1}{2}=\frac{1-\frac{3}{9}}{1+\left(\tan(30)\right)^{2}}$$

Whakahekea te hautanga \frac{3}{9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$\frac{1}{2}=\frac{1-\frac{1}{3}}{1+\left(\tan(30)\right)^{2}}$$

Tangohia te \frac{1}{3} i te 1, ka \frac{2}{3}.

$$\frac{1}{2}=\frac{\frac{2}{3}}{1+\left(\tan(30)\right)^{2}}$$

Get the value of \tan(30) from trigonometric values table.

$$\frac{1}{2}=\frac{\frac{2}{3}}{1+\left(\frac{\sqrt{3}}{3}\right)^{2}}$$

Kia whakarewa i te \frac{\sqrt{3}}{3} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$\frac{1}{2}=\frac{\frac{2}{3}}{1+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 1 ki te \frac{3^{2}}{3^{2}}.

$$\frac{1}{2}=\frac{\frac{2}{3}}{\frac{3^{2}}{3^{2}}+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

Tā te mea he rite te tauraro o \frac{3^{2}}{3^{2}} me \frac{\left(\sqrt{3}\right)^{2}}{3^{2}}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{1}{2}=\frac{\frac{2}{3}}{\frac{3^{2}+\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

Whakawehe \frac{2}{3} ki te \frac{3^{2}+\left(\sqrt{3}\right)^{2}}{3^{2}} mā te whakarea \frac{2}{3} ki te tau huripoki o \frac{3^{2}+\left(\sqrt{3}\right)^{2}}{3^{2}}.

$$\frac{1}{2}=\frac{2\times 3^{2}}{3\left(3^{2}+\left(\sqrt{3}\right)^{2}\right)}$$

Me whakakore tahi te 3 i te taurunga me te tauraro.

$$\frac{1}{2}=\frac{2\times 3}{\left(\sqrt{3}\right)^{2}+3^{2}}$$

Whakareatia te 2 ki te 3, ka 6.

$$\frac{1}{2}=\frac{6}{\left(\sqrt{3}\right)^{2}+3^{2}}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{1}{2}=\frac{6}{3+3^{2}}$$

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$\frac{1}{2}=\frac{6}{3+9}$$

Tāpirihia te 3 ki te 9, ka 12.

$$\frac{1}{2}=\frac{6}{12}$$

Whakahekea te hautanga \frac{6}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$\frac{1}{2}=\frac{1}{2}$$

Whakatauritea te \frac{1}{2} me te \frac{1}{2}.

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