Whakaoti i te {left(cos(45^circ)right)}^2-1/2tan(45^circ)+tan(30^circ)

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

Aromātai

$$\frac{\sqrt{3}}{3}\approx 0.577350269$$

Ngā Hipanga Rongoā Poto

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

$$\left(\frac{\sqrt{2}}{2}\right)^{2}-\frac{1}{2}\tan(45)+\tan(30)$$

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.

$$\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-\frac{1}{2}\tan(45)+\tan(30)$$

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

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

Whakareatia te \frac{1}{2} ki te 1, ka \frac{1}{2}.

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 2^{2} me 2 ko 4. Whakareatia \frac{1}{2} ki te \frac{2}{2}.

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

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

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

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

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 4 me 3 ko 12. Whakareatia \frac{\left(\sqrt{2}\right)^{2}-2}{4} ki te \frac{3}{3}. Whakareatia \frac{\sqrt{3}}{3} ki te \frac{4}{4}.

$$\frac{3\left(\left(\sqrt{2}\right)^{2}-2\right)}{12}+\frac{4\sqrt{3}}{12}$$

Tā te mea he rite te tauraro o \frac{3\left(\left(\sqrt{2}\right)^{2}-2\right)}{12} me \frac{4\sqrt{3}}{12}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{3\left(\left(\sqrt{2}\right)^{2}-2\right)+4\sqrt{3}}{12}$$

Ko te pūrua o \sqrt{2} ko 2.

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

Tangohia te 2 i te 2, ka 0.

$$\frac{0}{4}+\frac{\sqrt{3}}{3}$$

Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$0+\frac{\sqrt{3}}{3}$$

Ko te tau i tāpiria he kore ka hua koia tonu.

$$\frac{\sqrt{3}}{3}$$