Question

I have to solve this without using a calculator..HELP chick attached.http://bamboodock.wacom.com/doodler/0f7a3296-198f-4f30-a584-d41782b1e8e6

Answer 1

There's a handy dandy trig identity there. What does sinxcosx ALMOST look like?

Answer 2

1

Answer 3

Easier one:
\[\sin(2x) = 2\sin(x)\cos(x)\]

Answer 4

So when you multiply and divide all that crap on the right out, you have:
\[\sin(x)\cos(x) \approx 0.347\]

Answer 5

wait how did it become sin2x=2sinxcosx...that can happen?

Answer 6

It hasnt YET. But we can agree thats what you have so far, right?

Answer 7

(We havent applied the identity yet)

Answer 8

yes//

Answer 9

Ok, but we WANT it to look like 2sinxcosx...because thats what our trig identity is. So multiply both sides by 2, what do you have?

Answer 10

liek this??

Answer 11

Yep, go ahead and multiply and divide that crap on the right out so it doesnt make my face hurt. :P

Answer 12

okay. hold on, could you please check when i finish?

Answer 13

Yes.

Answer 14

Be sure that when you start rounding stuff off that you use the "squiggly equals" sign, since its an approximation. If you have a strict prof. they may strike a big red 'x' through it if you don't.

Answer 15

haha okay is this okay?

Answer 16

Perfect, now apply the trig identity.

Answer 17

\[2\sin(x)\cos(x) = \sin(2x)\]
This is the double angle sine identity.

Answer 18

oh okay!!

Answer 19

Take the arcsine of both sides and divide by two. You should get an answer close to what your graphing calc gave you on the last question.

Answer 20

Be sure your calculator is in degree mode if you want a degree measure answer.

Answer 21

ok...

Answer 22

Did it come out right, should be close to 22.01 or so.

Answer 23

yesss!! omgsh yes i got them!

Answer 24

Excellent. Just remember the trig identities and don't be afraid to manipulate an equation! You can make it look how you want as long as you follow the rules. Do to one side what you do to the other ;]

Answer 25

ok how do you remember all these identities? its so hard. on the tests were you allowed to look at the identites? or did you remeber them by memory?

Answer 26

Well, you can derive them from the fundamental identity:
\[\sin^2(x) + \cos^2(x) =1\]
and from other basic trig principles. However, if nobody has shown you, that may be tedious or difficult. I've been through alot of math, so much so that at this point it's just been beaten into my head and I'll never forget them. The more you practice the more you'll encounter them and you'll start remembering. Some are more useful than others.

Answer 27

okay because im taking online trigonometry and no ones teaching me so its so difficult.