152k views
3 votes
Find the value of all trigonometric functions of 135°

User Thyselius
by
8.5k points

1 Answer

6 votes

Answer:

sin (- 135°)= – sin 135°= – sin (1 × 90°+ 45°) = – cos 45° = – 1√2

cos (- 135°)= cos 135°= cos (1 × 90°+ 45°) = – sin 45°= – 1√2

tan (- 135°) = – tan 135° = – tan ( 1 × 90° + 45°) = – (- cot 45°) = 1

csc (- 135°)= – csc 135°= – csc (1 × 90°+ 45°)= – sec 45° = – √2

sec (- 135°)= sec 135°= sec (1 × 90°+ 45°)= – csc 45°= – √2

cot (- 135°) = – cot 135° = – cot ( 1 × 90° + 45°) = – (-tan 45°) = 1

Explanation:

hope this helps

User ParaMeterz
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories