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Is it possible $\sin\alpha=\cos\alpha=1?$

We have defined trigonometry using a unit circle. The point $X(1;1)$ does not lie on the trig circle, right? Does this mean that $\alpha$ does not exist?

kormoran
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    You are correct. – John Douma Dec 30 '20 at 15:53
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    it would contradict $\sin^2\alpha+\cos^2\alpha=1$ – J. W. Tanner Dec 30 '20 at 15:53
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    You can also see that it doesn't exist, thanks to the identity $\cos^2\alpha + \sin^2\alpha = 1$. If what you say above was possible, you would have $1=2$. –  Dec 30 '20 at 15:54
  • @DietrichBurde, I saw the question but I don't think I understand it. – kormoran Dec 30 '20 at 16:38
  • @J.W.Tanner, thank you for the response! I got it. But what if we have $\sin\alpha=\cos\alpha=\dfrac{\sqrt2}{2}?$ Does $\alpha$ exist then? I checked and the identity $\sin^2\alpha+\cos^2\alpha=1$ holds. Does this mean that $\alpha$ exist? Is this a sufficient condition for $\alpha$ to exist? – kormoran Dec 30 '20 at 16:39
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    Yes, $\sin(\pi/4)=\cos(\pi/4)=\dfrac{\sqrt2}2$, and there also exists $\theta$ such that $\sin(\theta)=\cos(\theta)=\dfrac{-\sqrt2}2$ – J. W. Tanner Dec 30 '20 at 16:51
  • Yes, but in case I don't know the exact values of the trig functions for the table angles, if I see that the mentioned identity holds, can I say that the angle exists? If $\sin^2\alpha+\cos^2\alpha=1$, then $\alpha$ exists for sure? – kormoran Dec 30 '20 at 16:52

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$\sin ^2 x + \cos ^2 x = 1$ for all $x$, so if you had $\cos \alpha = \sin \alpha = 1$, then $\sin ^2 \alpha + \cos ^2 \alpha$ would be equal to $2$. This is absurd.

math
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The answer you already gave is the clearest in my opinion. The trigonometric functions can be defined using the unit circle, so that $\sin(\theta)$ represents the $y$-coordinate when you have travelled $\theta$ degrees/radians anticlockwise around the unit circle, starting from the point $(1,0)$. Likewise, $\cos(\theta)$ represents the $x$-coordinate when you have travelled $\theta$ degrees/radians around.

Since the equation of a unit circle is $x^2+y^2=1$, and $(x,y)=(\cos(\theta),\sin(\theta))$, we obtain the identity $$ \cos(\theta)^2+\sin(\theta)^2=1 \, , $$ which math's answer references. This is one way to answer your question. More directly, we know that the point $(1,1)$ does not lie on the unit circle, since the distance from the origin to that point is $\sqrt{2}$. Hence, it is never true that $\sin(\theta)=\cos(\theta)=1$.

Joe
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  • Thank you for the response! Is the identity $\sin^2\alpha+\cos^2\alpha=1$ a sufficient condition for an angle to exist? – kormoran Dec 30 '20 at 16:51
  • @KaloyanK. Sine and cosine are defined for all values of $\alpha$, but it is always true that $\sin(\alpha)^2 + \cos(\alpha)^2=1$. Does that answer your question, or am I misunderstanding what you are saying? – Joe Dec 30 '20 at 16:55
  • No, you didn't understand what I meant. Let me try again. I mean can we say that if the identity holds, then the angle exists for sure? Are there any cases for which the identity holds but $\alpha$ does not exist... We saw that when $\sin\alpha=\cos\alpha=1$ the identity does not hold, so we conclduded that $\alpha$ does not exist. Can we do it the other way? – kormoran Dec 30 '20 at 16:57
  • @KaloyanK. It's difficult to understand what you mean, since you say 'if the identity holds...', but the identity always holds regardless of what value of $\alpha$ we are talking about. So $\alpha$ could be any one of infinitely many possible values. I am sorry if I am still misunderstanding you; feel free to clarify what you are saying further. – Joe Dec 30 '20 at 17:02
  • I'm sorry. English isn't my native language, so that doesn't help. :))But doesn't it hold only if $\alpha$ exists? – kormoran Dec 30 '20 at 17:03
  • @KaloyanK. What do you mean by if $\alpha$ exists? Could you phrase that in another way, please. – Joe Dec 30 '20 at 17:04
  • Well, now I am solving a problem: If $\sin\alpha=\cos\alpha=1,$ does $\alpha$ exist? Is there an angle $\alpha$ for which $\sin\alpha=\cos\alpha=1$ - maybe that's more clear. Does this help you understand? – kormoran Dec 30 '20 at 17:06
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    @KaloyanK. Ah, I think I understand. So $\alpha$ doesn't exist if $\sin(\alpha)=\cos(\alpha)=1$. (Although I would prefer saying 'there isn't a value of $\alpha$ such that $\sin(\alpha)=\cos(\alpha)=1$.) $\alpha$ does also not exist if $\sin(\alpha)^2+\cos(\alpha)^2 \neq 1$. Basically, if $x$ is any real number, then $\sin(x)^2+\cos(x)^2=1$. – Joe Dec 30 '20 at 17:10