# Thread: Find Exact Value Without Calculator

1. ## Find Exact Value Without Calculator

Let t = theta for short

For each question below, f(t) = sin(t) and g(t) = cos(t). Find the exact value of each function if t = 60 degrees. Do NOT use a calculator.

(1) g(t)

(2) [g(t)]^2

2. Are you really saying that you do not know $\cos (60^o)$?

3. ## no...

The question states that theta = 60 degrees

Can you help me?

4. ## no...

Originally Posted by Plato
Are you really saying that you do not know $\cos (60^o)$?
I know cos(60 degrees).

But what can I do to proceed?

5. $g(60^o)=\cos(60^o)$.
What do you know about functions?

6. To find the value of the trigonometric functions of the angles 30, 45, and 60, we use the special 45-45-90 and 30-60-90 triangles:
Special right triangles - Wikipedia, the free encyclopedia

You can also find values of an angle whose sum/difference is two of the mentioned angles using the sum/difference identities.

7. Originally Posted by Chop Suey
To find the value of the trigonometric functions of the
Your above response seems to say no indeed.
The poster said that there were no problem with the evaluation of $\cos(60^o)$.
Thus the problem must be with understanding the concept of function.

8. ## 1/2...

Originally Posted by Plato
$g(60^o)=\cos(60^o)$.
What do you know about functions?
Cos(60) = 1/2

Then g(t) = 1/2

Is this right?

============

If g(t) = 1/2, the (1/2)^2 = 1/4

Is this right?

9. Originally Posted by Plato
Your above response seems to say no indeed.
The poster said that there were no problem with the evaluation of $\cos(60^o)$.
Thus the problem must be with understanding the concept of function.
Plato: I read the entire thread before posting. I thought that the matter is how to find the values of trigonometric functions and that the poster was asking how to find them without using a calculator. The opposite prompt said she knows what the value is, but she left an implication that she doesn't know how to get to it. I know what she means now, sorry for the misunderstanding.

Magentarita: yes, you are right.

10. ## ok...

I'm glad to have been able to solve each question.

Tricky but not hard.