Whakaoti i te sin^{2}30^{circ}+sin60^{circ}-sin^245^circ+cos^230^circ

$$\sin ^ { 2 } 30 ^ { \circ } + \sin 60 ^ { \circ } - \sin ^ { 2 } 45 ^ { \circ } + \cos ^ { 2 } 30 ^ { \circ }$$

Aromātai

$$\frac{\sqrt{3}+1}{2}\approx 1.366025404$$

Ngā Hipanga Rongoā Poto

Tīkina te uara \sin(30) mai i te ripanga uara pākoki.

$$\left(\frac{1}{2}\right)^{2}+\sin(60)-\left(\sin(45)\right)^{2}+\left(\cos(30)\right)^{2}$$

Tātaihia te \frac{1}{2} mā te pū o 2, kia riro ko \frac{1}{4}.

$$\frac{1}{4}+\sin(60)-\left(\sin(45)\right)^{2}+\left(\cos(30)\right)^{2}$$

Tīkina te uara \sin(60) mai i te ripanga uara pākoki.

$$\frac{1}{4}+\frac{\sqrt{3}}{2}-\left(\sin(45)\right)^{2}+\left(\cos(30)\right)^{2}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 4 me 2 ko 4. Whakareatia \frac{\sqrt{3}}{2} ki te \frac{2}{2}.

$$\frac{1}{4}+\frac{2\sqrt{3}}{4}-\left(\sin(45)\right)^{2}+\left(\cos(30)\right)^{2}$$

Tā te mea he rite te tauraro o \frac{1}{4} me \frac{2\sqrt{3}}{4}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{1+2\sqrt{3}}{4}-\left(\sin(45)\right)^{2}+\left(\cos(30)\right)^{2}$$

Tīkina te uara \sin(45) mai i te ripanga uara pākoki.

$$\frac{1+2\sqrt{3}}{4}-\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(\cos(30)\right)^{2}$$

Kia whakarewa i te \frac{\sqrt{2}}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$\frac{1+2\sqrt{3}}{4}-\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}+\left(\cos(30)\right)^{2}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{1+2\sqrt{3}}{4}-\frac{2}{2^{2}}+\left(\cos(30)\right)^{2}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$\frac{1+2\sqrt{3}}{4}-\frac{2}{4}+\left(\cos(30)\right)^{2}$$

Tā te mea he rite te tauraro o \frac{1+2\sqrt{3}}{4} me \frac{2}{4}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{1+2\sqrt{3}-2}{4}+\left(\cos(30)\right)^{2}$$

Mahia ngā tātaitai i roto o 1+2\sqrt{3}-2.

$$\frac{-1+2\sqrt{3}}{4}+\left(\cos(30)\right)^{2}$$

Tīkina te uara \cos(30) mai i te ripanga uara pākoki.

$$\frac{-1+2\sqrt{3}}{4}+\left(\frac{\sqrt{3}}{2}\right)^{2}$$

Kia whakarewa i te \frac{\sqrt{3}}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$\frac{-1+2\sqrt{3}}{4}+\frac{\left(\sqrt{3}\right)^{2}}{2^{2}}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakarohaina te 2^{2}.

$$\frac{-1+2\sqrt{3}}{4}+\frac{\left(\sqrt{3}\right)^{2}}{4}$$

Tā te mea he rite te tauraro o \frac{-1+2\sqrt{3}}{4} me \frac{\left(\sqrt{3}\right)^{2}}{4}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{-1+2\sqrt{3}+\left(\sqrt{3}\right)^{2}}{4}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{-1+2\sqrt{3}}{4}+\frac{3}{2^{2}}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$\frac{-1+2\sqrt{3}}{4}+\frac{3}{4}$$

Tā te mea he rite te tauraro o \frac{-1+2\sqrt{3}}{4} me \frac{3}{4}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{-1+2\sqrt{3}+3}{4}$$

Mahia ngā tātaitai i roto o -1+2\sqrt{3}+3.

$$\frac{2+2\sqrt{3}}{4}$$