Calculate the approximate value I am supposed to calculate the approximate value of $$\cos 151^\circ$$ My idea was that I can divide it in the form: $$\cos 90^\circ+ 61^\circ= \cos \frac{\pi}{2}+ \left (\frac{\pi}{3} +\frac{\pi}{180} \right )$$ Then I use the derivation for cosx:
$$-\sin \frac{\pi}{2}\left ( \frac{\pi}{3} +\frac{\pi}{180} \right )=-\frac{61\pi}{180}+61^\circ$$
But I guess, it is not correct.
Can anyone help me?
 A: \begin{equation}
\cos\left(150^\circ\right)=\cos\left(90^\circ-\left(-60^\circ\right)\right)=\sin\left(-60^\circ\right)=-\sin\left(60^\circ\right)=-\frac{\sqrt3}2\\
\sin\left(150^\circ\right)=\sin\left(90^\circ-\left(-60^\circ\right)\right)=\cos\left(-60^\circ\right)=\cos\left(60^\circ\right)=\frac12
\end{equation}
Using the fact that $\cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right)$, we get
\begin{equation}
\cos\left(151^\circ\right)=\cos\left(150^\circ+1^\circ\right)=\cos\left(150^\circ\right)\cos\left(1^\circ\right)-\sin\left(150^\circ\right)\sin\left(1^\circ\right)\\
=-\left(\frac{\sqrt3}2\cos\left(1^\circ\right)+\frac12\sin\left(1^\circ\right)\right)
\end{equation}
Since
\begin{equation}
\sin\left(x\right)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\\
\cos\left(x\right)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\\
\end{equation}
We have 
\begin{equation}
\sin\left(1^\circ\right)=\sin\left(\frac\pi{180}\right)\approx\frac\pi{180}-\frac1{3!}\left(\frac\pi{180}\right)^3\\
\cos\left(1^\circ\right)=\cos\left(\frac\pi{180}\right)\approx1-\frac1{2!}\left(\frac\pi{180}\right)^2
\end{equation}
Plugging these approximate values into the above formula for $\cos\left(151^\circ\right)$ we get
\begin{equation}
\cos\left(151^\circ\right)\approx-\left(\frac{\sqrt3}2\left(1-\frac1{2!}\left(\frac\pi{180}\right)^2\right)+\frac12\left(\frac\pi{180}-\frac1{3!}\left(\frac\pi{180}\right)^3\right)\right)\\
\approx-0.8746
\end{equation}
A: We have
$$\cos 151 = \cos(180-29)=-\cos (29)$$
and
$$\cos (29)=\cos \left(\frac \pi 6-\frac{\pi}{180}\right)=\\=\cos \left(\frac \pi 6\right)\cos\left(\frac{\pi}{180}\right)+\sin \left(\frac \pi 6\right)\sin\left(\frac{\pi}{180}\right)$$
then use that for $x\approx 0$ in radians


*

*$\cos x \approx 1$

*$\sin x \approx x$
to obtain
$$\cos (29)\approx\frac{\sqrt 3}2\cdot 1+\frac12 \cdot \frac{\pi}{180}\approx0.874752\ldots$$
A: One way is to use derivatives.
Let $y=f(x) = \cos x$
$dy = -\sin x dx$
Let $x=150^\circ = \frac{2\pi}{3}  , dx = 1^\circ\approx 0.0174$ 
$\cos x = \cos150^\circ \approx -0.8660, \sin x = \sin 150^\circ =0.5$ 
$y+dy = f(x+dx) = \cos x -\sin x dx = \cos(150^\circ)-\sin(150^\circ)\times0.0174 \approx-0.8747$
$\implies\cos(151^\circ) \approx -0.8747$
