Diketahui Alfa(a)=45° dan Beta(b)= 60°,Hitung!1. 2 × sin 45° × cos 60°2. sin 45° × cos 60° + sin 60°.... Question from @JohanIrawanto

February 2019 1 15 Report

Diketahui Alfa(a)=45° dan Beta(b)= 60°,Hitung!
1. 2 × sin 45° × cos 60°
2. sin 45° × cos 60° + sin 60° ×
cos 45 °
3. sin 45° × cos 60° - sin 60° × cos 45°
4. tan 45° + tan 60° /1(tan 45° × tan 60°)
5. sin²45° + cos²60° +sin²60° + cos²45°

kalau soal tidak jelas silakan lihat foto,bantuin Aku yg masih junior kak

idznizhahrina Kelas : 10
Mapel : Matematika
Kategori : Trigonometri Dasar
Kata kunci : trigonometri sudut istimewa kuadran 1
Kode : 10.2.6 (Kelas 10 Matematika Bab 6-Trigonometri Dasar)

Pembahasan:

untuk mengerjakan soal trigonometri, kita harus hafal dahulu nilai sin cos tan untuk sudut-sudut istimewa di kuadran 1 seperti berikut ini:

$\sin 0^{\circ}=0 \\ \sin 30^{\circ}= \frac{1}{2} \\ \sin 45^{\circ}= \frac{1}{2} \sqrt{2} \\ \sin 60^{\circ}= \frac{1}{2} \sqrt{3} \\ \sin 90^{\circ}=1 \\ \\ \cos 0^{\circ}=1 \\ \cos 30^{\circ}= \frac{1}{2} \sqrt{3} \\ \cos 45^{\circ}= \frac{1}{2} \sqrt{2} \\ \cos 60^{\circ}= \frac{1}{2} \\ \cos 90^{\circ}=0$

$\tan 0^{\circ}=0 \\ \tan 30^{\circ}= \frac{1}{3} \sqrt{3} \\ \tan 45^{\circ}= 1 \\ \tan 60^{\circ}= \sqrt{3}$
$\tan 90^{\circ}=$ tak terdefinisi

$A.\; 2\times \sin 45^{\circ}\times \cos 60^{\circ} \\ =2\times \frac{1}{2} \sqrt{2}\times \frac{1}{2} \\ = \frac{1}{2} \sqrt{2}$

$B.\; \sin 45^{\circ}\cos 60^{\circ}+\sin 60^{\circ}\cos 45^{\circ} \\ = \frac{1}{2} \sqrt{2}\times \frac{1}{2}+ \frac{1}{2} \sqrt{3}\times \frac{1}{2} \sqrt{2} \\ = \frac{1}{4} \sqrt{2}+ \frac{1}{4} \sqrt{6} \\ = \frac{1}{4} (\sqrt{2}+ \sqrt{6})$

$C.\; \sin 45^{\circ}\cos 60^{\circ}-\sin 60^{\circ}\cos 45^{\circ} \\ = \frac{1}{2} \sqrt{2}\times \frac{1}{2}- \frac{1}{2} \sqrt{3}\times \frac{1}{2} \sqrt{2} \\ = \frac{1}{4} \sqrt{2}- \frac{1}{4} \sqrt{6} \\ = \frac{1}{4} (\sqrt{2}- \sqrt{6})$

$D.\; \frac{\tan45^{\circ} +\tan 60^{\circ}}{1-(\tan 45^{\circ} \tan 60^{\circ})} \\ = \frac{1+ \sqrt{3} }{1- \sqrt{3} }\times \frac{1+ \sqrt{3} }{1+ \sqrt{3} } \\ = \frac{1+ \sqrt{3}+ \sqrt{3} +3 }{1+ \sqrt{3}- \sqrt{3}-3 } \\ = \frac{4+2 \sqrt{3} }{-2} \\ =-2- \sqrt{3}$

$E.\; \sin^2 45^{\circ}+\cos^260^{\circ}+\sin^260^{\circ}+cos^245^{\circ} \\ =( \frac{1}{2} \sqrt{2} )^2+( \frac{1}{2} )^2+( \frac{1}{2} \sqrt{3} )^2+( \frac{1}{2} \sqrt{2} )^2 \\ = \frac{2}{4}+ \frac{1}{4}+ \frac{3}{4}+ \frac{2}{4} \\ = \frac{8}{4} \\ =2$

soal lainnya tentang trigonometri sudut istimewa di kuadran 1 yang dapat dipelajari:
brainly.co.id/tugas/6329482

Semangat belajar!
Semoga membantu :)

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