Write $\cos(x)$ in terms of $\sin(x)$ if the terminal point $x$ is in quadrant IV.
I know $\cos^2(x)$ $= 1-\sin^2(x)$. And I know that cos is positive in quadrant IV. I am
guessing that the answer is $\sqrt{1-\sin^2(x)}$ Can anyone help verify this answer?
Thank you!!
$\endgroup$ 12 Answers
$\begingroup$As written, it's not quite correct. Let's start with the identity
$$\sin^2{x} + \cos^2{x} = 1 \implies \cos^2{x} = 1 - \sin^2{x}$$
In quadrant IV, $\cos{x}$ is positive, so taking a square root gives us
$$\cos{x} = \sqrt{1 - \sin^2{x}}$$
$\endgroup$ $\begingroup$$\cos x = 1 - 2 * \sin^2(x/2)$
prove:
\begin{align}
&=>1 - 2 * \sin^2(x/2) \\
&= 1 - 2 * \sin(x/2) * \sin(x/2) \\
&= 1 - 2 * \sqrt{((1-\cos x)/2)} * \sqrt{((1-\cos x)/2)} \quad \text{NOT: even sqrts are (-), (-)*(-)=(+)} \\
&= 1 - 2 * (1-\cos x)/2 \\
&= 1 - 1 + \cos x \\
&= \cos x \\
\end{align}
