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ฟังก์ชันตรีโกณมิติของผลบวกและผลต่างของมุม 26997

Authored by Naraphat Viriyolan

Mathematics

11th Grade

Used 9+ times

ฟังก์ชันตรีโกณมิติของผลบวกและผลต่างของมุม 26997
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20 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

จงหาค่าของ sin 75°\sin\ 75\degree   

 64+24\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}  

0

 6424\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}  

 32\frac{\sqrt{3}}{2}  

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

จงหาค่าของ  sin 15°\sin\ 15\degree  

 64+24\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}  

1

0

 6424\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}  

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

จงหาค่าของ cos 75°\cos\ 75\degree  

 32\frac{\sqrt{3}}{2}  

 6424\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}  

0

 32\frac{-3}{2}  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

จงหาค่าของ  cos 15°\cos\ 15\degree  

 72\frac{\sqrt{7}}{2}  

0

 64+24\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}  

 32\frac{\sqrt{3}}{2}  

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

กำหนดให้ 0<A<π20<A<\frac{\pi}{2}   และ  0<B<π20<B<\frac{\pi}{2}  ถ้า  sin A=35\sin\ A=\frac{3}{5}  และ  cos B=513\cos\ B=\frac{5}{13}  จงหา

 5665\frac{56}{65}  

 6365\frac{63}{65}  

 1665\frac{-16}{65}  

 3365\frac{-33}{65}  

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

กำหนดให้ π2<A<π\frac{\pi}{2}<A<\pi   และ  0<B<π20<B<\frac{\pi}{2}  ถ้า  cos A =817\cos\ A\ =-\frac{8}{17}  และ  sin B =1213\sin\ B\ =\frac{12}{13}  จงหา  cos(AB)\cos\left(A-B\right)  

 42140\frac{\sqrt{42}}{140}  

 40221\frac{-40}{221}  

 180221\frac{180}{221}  

 140221\frac{140}{221}  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

กำหนดให้  π2<A<π\frac{\pi}{2}<A<\pi  และ  π2<B<π\frac{\pi}{2}<B<\pi  ถ้า  sin A=35\sin\ A=\frac{3}{5}  และ  cos B =513\cos\ B\ =-\frac{5}{13}  แล้วจงหา  cos(A+B)\cos\left(A+B\right)  

 3665-\frac{36}{65}  

 2065\frac{20}{65}  

 1665\frac{-16}{65}  

 5665\frac{56}{65}  

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