If cos 'theta' =1/2 the sin'theta' = WHAT?

[rere]

Help please...

If 90 degrees < 'theta' <180 degrees what is equaled to 5 cos 'theta' sec 'theta'?

and another...

If cos 'theta' =1/2 the sin'theta' = WHAT?

thank you for any advice and help.

[stapel_eliz]

If 90 degrees < 'theta' <180 degrees what is equaled to 5 cos 'theta' sec 'theta'?

I'm sorry, but I don't understand what this exercise is asking...? Are you supposed to find a range of value for the product of the 5, the cosine, and the secant? Or something else?

If cos 'theta' =1/2 the sin'theta' = WHAT?

Draw the right triangle. Label "adjacent" and "hypotenuse" sides with the values from the cosine ratio. Use the Pythagorean Theorem to find the value for the "opposite" side. Read off the value of the sine from the picture.

[rere]

Yes, I am not sure what the question is asking for, I wrote it just as the problem said...Perhaps it is a typo
On the other, your advice helped me a lot. Thanks.
How about this problem;
If f9x0=x^2-x, which of the following is equal to x^2+x
A. f(x)+x
B. f(x)-x
C. -f(x)
D. f(-x)
E. [f(x)]^2
Thanks

[stapel_eliz]

If f9x0=x^2-x, which of the following is equal to x^2+x

I will guess that the first bit is supposed to be

If so, try plugging into the various expressions, in places of , and find the one that matches.

[pinkpuppi]

the square root of three over 2. remember SOH CAH TOA. if you make a triangle with 'theta' as your angle then plug in the cos in the appropriate spots. 1 in the adjacent leg and 2 on the hypotonuse. this would create a 30 60 90 triangle, meaning the only available side should be root three. then take that over two for sin because that would be the opposit angle. hope that makes sense!