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1 year ago

Find the value of all trigonometric functions of 135°

Mathematics

1 answer:

Vera_Pavlovna [14]1 year ago

5 0

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

Step-by-step explanation:

hope this helps

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Answer:

Step-by-step explanation:

<u>Equations</u>

Let's call

b= base of the triangle

h=height of the triangle

The base is 9 cm longer than the height, thus:

b = h + 9

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