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Trigonometri Kls XI

Authored by Ratna Sari 72

Mathematics

12th Grade

CCSS covered

Used 39+ times

Trigonometri Kls XI
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15 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 sin(105°)=...\sin\left(105\degree\right)=...  

 \frac{1}{4}\sqrt{8}  

 12(2+6)\frac{1}{2}\left(\sqrt{2}+\sqrt{6}\right)  

 14(2+6)\frac{1}{4}\left(\sqrt{2}+\sqrt{6}\right)  

 14(26)\frac{1}{4}\left(\sqrt{2}-\sqrt{6}\right)  

 14(62)\frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)  

Tags

CCSS.HSF.TF.C.9

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 cos(75°)=...\cos\left(75\degree\right)=...  

 \frac{1}{4}\sqrt{8}  

 12(2+6)\frac{1}{2}\left(\sqrt{2}+\sqrt{6}\right)  

 14(2+6)\frac{1}{4}\left(\sqrt{2}+\sqrt{6}\right)  

 14(26)\frac{1}{4}\left(\sqrt{2}-\sqrt{6}\right)  

 14(62)\frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)  

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 cos85°cos55°+sin85°sin55°=...\cos85\degree\cos55\degree+\sin85\degree\sin55\degree=...  

0

1

 12\frac{1}{2}  

 122\frac{1}{2}\sqrt{2}  

 123\frac{1}{2}\sqrt{3}  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 sin75°cos15°cos75°sin15°=...\sin75\degree\cos15\degree-\cos75\degree\sin15\degree=...  

0

1

 12\frac{1}{2}  

 122\frac{1}{2}\sqrt{2}  

 123\frac{1}{2}\sqrt{3}  

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Nilai dari  tan75°\tan75\degree  adalah...

 2+3-2+\sqrt{3}  

 23-2-\sqrt{3}  

 2+32+\sqrt{3}  

 232-\sqrt{3}  

 3+23+\sqrt{2}  

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Rumus sinus jumlah dua sudut...

sin (α+β)=sin α cos β+cos α sin β\sin\ \left(\alpha+\beta\right)=\sin\ \alpha\ \cos\ \beta+\cos\ \alpha\ \sin\ \beta

sin (αβ)=sinα cos βcos α sin β\sin\ \left(\alpha-\beta\right)=\sin\alpha\ \cos\ \beta-\cos\ \alpha\ \sin\ \beta

sin (α+β)=sin α cos βcos α sin β\sin\ \left(\alpha+\beta\right)=\sin\ \alpha\ \cos\ \beta-\cos\ \alpha\ \sin\ \beta

sin (αβ)=sin α cos β+cos α sin β\sin\ \left(\alpha-\beta\right)=\sin\ \alpha\ \cos\ \beta+\cos\ \alpha\ \sin\ \beta

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Rumus tangen jumlah dua sudut...

tan (α+β)=tan αtan β1+tan α tan β\tan\ \left(\alpha+\beta\right)=\frac{\tan\ \alpha-\tan\ \beta}{1+\tan\ \alpha\ \tan\ \beta}

tan (α+β)=tanβ+tan α 1+tan α tan β\tan\ \left(\alpha+\beta\right)=\frac{\tan\beta+\tan\ \alpha\ }{1+\tan\ \alpha\ \tan\ \beta}

tan (α+β)=tan α+tan β1tan α tan β\tan\ \left(\alpha+\beta\right)=\frac{\tan\ \alpha+\tan\ \beta}{1-\tan\ \alpha\ \tan\ \beta}

tan (αβ)=tan αtan β1+tan α tan β\tan\ \left(\alpha-\beta\right)=\frac{\tan\ \alpha-\tan\ \beta}{1+\tan\ \alpha\ \tan\ \beta}

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