Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$. Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$ Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$.Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$
This is an analysis question and I can only use the fundamental definition of $\sin$ and $\cos$ that is the infinite sum. I can also use the Taylor theorem, but I am stuck and clearly do not understand what I need to assume and what I need to prove. I can assume that $\sin$ and $\cos$ are infinitely differentiable. $\pi$ is defined as$:\quad\pi= 2\inf\{x\geq 0:\quad cos x= 0\}$.
Intuitively I should be using the Intermediate Value Theorem but I am not sure on which function.
So all you have is the power series for $\sin x$ and $\cos x$, and you know that $\frac \pi2$ is the smallest positive zero of $\cos x$.
Lemma. Let $f\colon\Bbb R\to\Bbb R$ be a twice differentiable function with
$$ f''(t)+f(t)=0\text{ for all }t,\quad f(0)=0,\quad f'(0)=0. $$
Then $f$ is identically $0$.
Proof. Indeed,
$$ \frac{\mathrm d}{\mathrm dx}(f(x)^2+f'(x)^2)
=2f'(x)f(x)+2f''(x)f'(x)=2f'(x)(f(x)+f''(x))=0$$
so that $f(x)^2+f'(x)^2$ is constant, and by evaluating at $0$, is constant zero.  But as squares of reals are non-negative, this implies that also $f$ is identically zero. $\square$
From the power series, you readily find that $$\sin't=\cos t,\qquad\cos't=-\sin t,$$ hence $$\sin''t=-\sin t,\qquad \cos''t=-\cos t.$$
As $\cos0=1$ and $\cos\frac\pi2=0$, the Intermediate Value Theorem gives us some $a\in[0,\frac\pi2]$ where $\cos a=|x|$ and then necessarily $|\sin a|=y$.
Note that $\sin t&gt;0$ for  $t\in(0,\frac\pi2]$ (and in particular, $\sin\frac\pi2=+1$) as otherwise Rolle's theorem would give us a critical point of $\sin t$ and hence a zero of $\cos t$  in $(0,\frac\pi2)$, contradicting the definition of $\pi$. We conclude that in fact $\sin a=y$. Thus
Result 1. If $x^2+y^2=1$, $x\ge 0$, $y\ge 0$, then there exists $a\in[0,\frac\pi2]$ with $\sin a=y$, $\cos a=x$.
By switching some roles, this also gives us
Result 2. If $x^2+y^2=1$, $x\le 0$, $y\ge 0$, then there exists $b\in[0,\frac\pi2]$ with $\sin b=-x$, $\cos b=y$.
If the $x$ in our original problem is $\ge0$, result 1 gives us the desired answer. So assume $x&lt;0$ and let $b$ be as in result 2.
Define $f(t)=\sin t+\cos(t+\frac\pi2)$. Then

$f(0)=\sin 0+\cos\frac\pi2=0+0=0$
$f'(0)=\cos 0-\sin\frac\pi2=1-1=0$
$f''(t)=-\sin t-\cos(t+\frac\pi2)=-f(t)$

Hence by the lemma, $f(t)=0$ for all $t$, i.e., $$\cos(t+\tfrac\pi2)=-\sin t$$
and similarly
$$\sin(t+\tfrac\pi2)=\cos t$$ for all $t\in\Bbb R$. In particular, with $c:=b+\frac\pi 2\in [0,\pi]$, we have
$$ \sin c = \cos b = y,\qquad \cos c = -\sin b = x$$
as desired.
$y=+\sqrt{1-x^2}&gt;0.$
Let $x=\cos t ,t\in[0,\pi] \implies y=\sin t &gt;0$