Question

Trig Question - Solve: sin^2 (theta) + cos (theta) =2 (Hint: Use the Pythagorean identity, sin^2 (theta) +cos^2 (theta) = 1 , to replace sin^2(theta) in the given equation).

I still don't quite know what to do or how to show work, please help!

Answer 1

Right. So if sin^2(x) + cos^2(x) = 1, you could also say either sin^2(x) is the same as 1 - cos^2(x) or that cos^2(x) is the same as 1-sin^2(x). That's what theyre hinting at you to do.

Answer 2

so are they asking me to find theta?

Answer 3

Nah. Just that you can do this:

Answer 4

so would sin^2 =1 - cos^2(x) be my answer?

Answer 5

I think they do want you to find theta.

Answer 6

Well eventually, yes, but just that he can turn sin^2(x) into 1-cos^2(x), allowing him to do another quadratic solving thing to eventually get theta.

Answer 7

Sorry if I made it confusing.

Answer 8

I don't think so...
\[\left( 1-\cos^2\theta \right)+\cos \theta=2\]
Is where you are no go from there.

Answer 9

I'm not sure how to find theta from that

Answer 10

Well I would substitute a variable say u in place of cos then simplify. So what would you have if you subbed u in for cos?

Answer 11

(1-u^2 θ) + u θ =2 ? then factor?

Answer 12

Well close but remember theta is the argument of cos so you really arent replacing cos but cos theta. so (1-u^2) + u = 2

Answer 13

Are you sure that the original equation was \[\sin^2 \theta + \cos \theta = 2\]?

Answer 14

I dont think it was, because there is no solution if that is the original.

Answer 15

This is what it looks like

Answer 16

look in the top left corner

Answer 17

Yep because the greatest value of cos theta will be 1 and that only occurs when theta is ... 0, 2pi, 4pi,... and the max value of sin^2 theta is 1 as well and that occurs when theta is pi/2, 3pi/2, 5pi/2 ... so these two values will never add to two so the final answer is no solution. If it is as written.

Answer 18

\[\sin^2 (\theta) + \cos (\theta) =2\]
\[1-\cos^2(\theta )+\cos(\theta)=2\]
\[-\cos^2(\theta)+\cos(\theta)=1\]
\[\cos^2(\theta)-\cos(\theta)+1=0\]

Answer 19

Unless satellite has some magic, I say no solution, too.

Answer 20

solve
\[x^2-x+1=0\] using the quadratic formula

Answer 21

then take the inverse cosine of the result

Answer 22

That would yield a complex number so in there is no solution in the real numbers.

Answer 23

@satellite73

Answer 24

lol so it will!

Answer 25

Haha, damn, no magic.

Answer 26

yeah no solutions

Answer 27

in fact you know there are no solutions from the start

Answer 28

No solutions in the Real numbers

Answer 29

you have
\[\sin^2(\theta)+\cos(\theta)=2\]

Answer 30

That is what I said! see above

Answer 31

the largest \(\cos(\theta)\) can be is 1

Answer 32

Yes as i said above.

Answer 33

and also the largest \(\sin^2(\theta)\) can be is 1
and they are not one at the same place

Answer 34

yes as I said above.

Answer 35

Probably the reason they gave you the pythagorean hint, just so you could see that you're going to find a way to get 2 out of that.

Answer 36

yeah so i see
i am a little slow

Answer 37

*not going to

Answer 38

Thanks for the support. :-)

Answer 39

yeah i was wondering why they gave me the Pythagorean identity

Answer 40

well thank you all so so much for helping me! :)

Answer 41

I think it was a poorly designed problem.

Answer 42

A very implicit way to prove a point I guess :/

Answer 43

yeah i totally agree

Answer 44

Back to the math cave.