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Korolek [52]
3 years ago
12

Evaluate (if possible) the six trigonometric functions of the real number. (If not possible, enter IMPOSSIBLE.)

Mathematics
1 answer:
Lady_Fox [76]3 years ago
4 0
T = 3π/2

Let's transform 3π/2 on degree

We know, π = 180° on trigonometric

Then,

3π / 2 = 3.(180°)/2

3π / 2 = 270°

Now let's evaluate.

Sin(270°) = Sin(-90°)

As Sin(-x) = - Sin(x)

Then,

= - Sin(90°)

= - 1

It is possible
_______________

2 example:

Csc(270°) = ?

As Csc(x) = 1 / Sin(x)

Then,

Csc(270°) = 1 / Sin(270°)

= 1 / - 1

= -1

It is possible
_____________

3 example:

Cos(270°) = Cos(-90°)

As Cos( -x) = Cos(x)

Then,

= Cos(90°)

= 0
_______________

4 example:

Sec(270°) = ?

As Sec(x) = 1 / Cos(x)

Then,

Sec(270°) = 1 / Cos(270°)

= 1 / 0

It isn't possible division by zero
______________

5 example

Tg(270°) = ?

As Tg(x) = Sin(x) / Cos(x)

Then,

Tg(270°) = Sin(270°) / Cos(270°)

= - 1 / 0

It isn't possible too.
______________

Now the six and last example

Cot (270°) = ?

As Cot(x) = 1 / tg(x)

Cotg (x) = Cos(x) / Sin(x)

Then,

Cot(270°) = Cos(270°) / Sin(270°)

= 0 / -1

= 0

It is possible
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