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

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

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...

Whats that about 3+4=5??

Is your question: prove that $sin x + cos x \neq 1$?

If so, $x = 2 \pi n$ or $\displaystyle x = 2\pi n + \frac{\pi}{2}$ for any $n \in \mathbb{Z}$ is an easy counterexample.

Whats that about 3+4=5??

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

Just sub in a random value for x which doesn't make either one 0 or 1.

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

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.

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.

I am still lost ...




I am sure @Zacken will help here

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.

Neither of you make any sense.

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

Perfect. Thank you.



lol?

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

Did you mean sin^2x + cos^2 x

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.

$x = 2\pi$.

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?