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# Prove sin(x) + cos(x) /=/ 1 watch

1. I get the whole 3+4 /=/ 5 argument

but from sin^2(x) + cos^2(x) = 1

how do we establish a proof to show that it is incorrcect

EDIT - I made a mess of this question. What I meant to say is that:

sin^2(x) + cos^2(x) = 1 for all values of x

However

sin(x) + cos(x) /=/ 1 for all values of x.

Prove why this is not the case.
2. (Original post by Mihael_Keehl)
I get the whole 3+4 /=/ 5 argument

but from sin^2(x) + cos^2(x) = 1

how do we establish a proof to show that it is incorrcect
you are not making sense...
3. (Original post by Mihael_Keehl)
I get the whole 3+4 /=/ 5 argument

but from sin^2(x) + cos^2(x) = 1

how do we establish a proof to show that it is incorrcect
4. (Original post by Mihael_Keehl)
I get the whole 3+4 /=/ 5 argument

but from sin^2(x) + cos^2(x) = 1

how do we establish a proof to show that it is incorrcect
Is your question: prove that ?

If so, or for any is an easy counterexample.
5. (Original post by TeeEm)
you are not making sense...
(Original post by raman_17)
well for 3^2 + 4^2 = 5^2 square root everything.

(Original post by Zacken)
Is your question: prove that ?

If so, or for any is an easy counterexample.
I see.

Nothing we can do by induction or anytihng else.
6. Just sub in a random value for x which doesn't make either one 0 or 1.
7. (Original post by raman_17)
8. just substitute a random number in and prove by counter example.
9. (Original post by Mihael_Keehl)
I get the whole 3+4 /=/ 5 argument

but from sin^2(x) + cos^2(x) = 1

how do we establish a proof to show that it is incorrcect
Remember sin(X) = opposite/hypotenuse and cos(X) = adjacent/hypotenuse

So you can see why the sum of squares is 1 by Pythagoras' theorem, but for the sum itself it can never be 1 (unless X=0).

Sin(X) + cos(X) = (O + A)/H.

If you think of the hypotenuse as the shortest distance between two points, A+O must be more than H.
10. (Original post by Vikingninja)
Just sub in a random value for x which doesn't make either one 0 or 1.
(Original post by Mihael_Keehl)
well for 3^2 + 4^2 = 5^2 square root everything.
Neither of you make any sense.

1. sin x + cos x DOES EQUAL 1. You can't prove it doesn't because it does.
11. I am still lost ...

I am sure Zacken will help here
12. (Original post by cubic^3)
just substitute a random number in and prove by counter example.
Remember sin(X) = opposite/hypotenuse and cos(X) = adjacent/hypotenuse

So you can see why the sum of squares is 1 by Pythagoras' theorem, but for the sum itself it can never be 1.
...
13. (Original post by Mihael_Keehl)
I get the whole 3+4 /=/ 5 argument

but from sin^2(x) + cos^2(x) = 1

how do we establish a proof to show that it is incorrcect
Add 2sinxcosx to both sides, you get sin^2(x) + 2sinxcosx + cos^2(x) = 1 + 2sinxcosx

This can be rewritten as (sinx + cosx)^2 = 1 + 2sinxcosx

This implies that sinx + cosx = ±sqrt(1 + sin2x), a solution does not exist for any value of x. Maybe you could even graph the function.

Edit: nah, I was wrong.
14. (Original post by aymanzayedmannan)
Add 2sinxcosx to both sides, you get sin^2(x) + 2sinxcosx + cos^2(x) = 1 + 2sinxcosx

This can be rewritten as (sinx + cosx)^2 = 1 + 2sinxcosx

This implies that sinx + cosx = ±sqrt(1 + sin2x), a solution does not exist for any value of x. Maybe you could even graph the function.
.
15. (Original post by TeeEm)
I am still lost ...

I am sure Zacken will help here
I've got no clue what's going on here either.
Remember sin(X) = opposite/hypotenuse and cos(X) = adjacent/hypotenuse

So you can see why the sum of squares is 1 by Pythagoras' theorem, but for the sum itself it can never be 1.

Sin(X) + cos(X) = (O + A)/H.

If you think of the hypotenuse as the shortest distance between to points, A+O must be more than H.
Perfect. Thank you.

(Original post by Zacken)
Neither of you make any sense.

1. sin x + cos x DOES EQUAL 1. You can't prove it doesn't because it does.
lol?

if x=30 then that yields 0.5(1 + sqrt3)

Did you mean sin^2x + cos^2 x
17. (Original post by Mihael_Keehl)
Perfect. Thank you.

lol?

if x=30 then that yields 0.5(1 + sqrt3)

Did you mean sin^2x + cos^2 x
No, your question makes 0 sense.
18. (Original post by aymanzayedmannan)
Add 2sinxcosx to both sides, you get sin^2(x) + 2sinxcosx + cos^2(x) = 1 + 2sinxcosx

This can be rewritten as (sinx + cosx)^2 = 1 + 2sinxcosx

This implies that sinx + cosx = ±sqrt(1 + sin2x), a solution does not exist for any value of x. Maybe you could even graph the function.
Another good way, thank you - I hadn't thought of doing this.

Any reason why you chose 2sinxcosx?

Any other value could be added right?
19. (Original post by Zacken)
.
True but that's not really a triangle is it. But I agree stupid question unless you exclude 0 and 2pi from the domain.
20. (Original post by Mihael_Keehl)
Another good way, thank you - I hadn't thought of doing this.

Any reason why you chose 2sinxcosx?

Any other value could be added right?
No it's the non squared part of (sinx+cosx)^2

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