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

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True but that's not really a triangle is it. But I agree stupid question unless you exclude 0 and 2pi from the domain.

x = 2pi, x = 4pi, x = 6pi, x = 8pi, x = 10pi, x = 12pi, x =14pi, ... I could literally go on forever.

The question does not specify anything about a range for x or it being in a triangle. Trigonometric functions are not by default restrained to triangles.

Reply 21

Mihael_Keehl

Original post by Zacken

No, your question makes 0 sense.

Ayman and asklepios answered what I was looking for, sorry if I did not construe the question well enough :P

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.

Shoudln't both sin and cos be squared here?

Reply 22

Zacken

Original post by Mihael_Keehl

Ayman and asklepios answered what I was looking for, sorry if I did not construe the question well enough :P

Shoudln't both sin and cos be squared here?

Ayman was wrong and he admitted it himself, read his edited post.

sin x + cos x does equal 1 for certain values of x.

Reply 23

Asklepios

Original post by Zacken

I'll go back to the biology forums

Reply 24

Ayman!

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?

It's just a bit of algebra to get it in the form a^2 + 2ab + b^2, but it solutions do exist for 2pi, and other periodic values.

Reply 25

Zacken

Original post by Asklepios

I'll go back to the biology forums

In fact, you have sin x and cos x defined over the whole complex plane.

Reply 26

Kvothe the Arcane

Original post by Mihael_Keehl

Shoudln't both sin and cos be squared here?

cosx+sinx=1 is true when x is a multiple of 2pi.

It's a pointless question because k2pi is literally the answer to finding x.

It's like saying x=7, prove that x is not not 7 or something.

Reply 27

Zacken

Original post by Kvothe the arcane

cosx+sinx=1 is true when x is a multiple of 2pi.

It's a pointless question because k2pi is literally the answer to finding x.

Not just that, x = 2pi k + pi/2 works as well.

Reply 28

Ayman!

Original post by Mihael_Keehl

Ayman and asklepios answered what I was looking for, sorry if I did not construe the question well enough :P

Shoudln't both sin and cos be squared here?

Nah mate, I actually proved it otherwise. All periodic values yield a solution.

Reply 29

tazarooni89

What if x is zero?
Am I missing something?

Reply 30

Zacken

Original post by tazarooni89

What if x is zero?
Am I missing something?

Nopes, the OP and several others are.

x can be any multiple of 2pi or x = 2pi k + pi/2 for any integer k.

Reply 31

XxKingSniprxX

Original post by raman_17

Whats that about 3+4=5??

I think she is referring to a 3, 4, 5 triangle perhaps. :s-smilie:

Reply 32

raman_17

Original post by XxKingSniprxX

I think she is referring to a 3, 4, 5 triangle perhaps. :s-smilie:

cheers for that :biggrin:

Original post by Mihael_Keehl

well for 3^2 + 4^2 = 5^2 square root everything.

I see.

So just by contradiction then?

Nothing we can do by induction or anytihng else.

nope........

Im late... Oh well

Original post by Zacken

ffs I wrote it misleadingly in the OP by accident. I was confused what you were saying which is probably in part due to the inarticulatirity of how I wrote the question.

I understand what you were saying now, could you please refer to the edited OP.

Original post by aymanzayedmannan

It's just a bit of algebra to get it in the form a^2 + 2ab + b^2, but it solutions do exist for 2pi, and other periodic values.

Ah yes, I see it know.

Original post by XxKingSniprxX

I think she is referring to a 3, 4, 5 triangle perhaps. :s-smilie:

Yes I was

Reply 37

Ayman!

Original post by Mihael_Keehl

I'll just refer to my initial "proof" to show you why this is not the case. The range of sin2x falls between -1 and 1. This implies that sinx + cosx can take up any real value between -sqrt(2) and sqrt(2) so a solution of 1 must exist.

Reply 38

Mihael_Keehl

Original post by aymanzayedmannan

Thanks yes, I see :biggrin:

Reply 39

Argylesocksrox

Ok, given you're excluding certain domains, you may write sinx+cosx as (root2)sin(x+pi/4), which means sin(x+pi/4)=1/root2, the inverse of such will give answers pi/4 or 3pi/4, which means x=pi/2 or 0, given I presume these values are excluded in the domain then there is no solution for sinx+cosx=1

I can't be bothered to write it more rigorously but that's what you need.