How do you do trig decimals without a calculator like $\cos \left( 5.22 \right)$?

Hint: For solving the trigonometric decimals given as $\cos \left( 5.22 \right)$, we have to use the series expression of the cosine and the sine functions given respectively as $\cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-......$ and $\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-......$. We first need to simplify the argument as the largest multiple of $\dfrac{\pi }{2}$. For this, we can use the approximate value of $\pi $ which is $3.14$. Then using the identity $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ we will get the final answer.

Complete step by step solution:
Let us write the given trigonometric expression as
$\Rightarrow E=\cos \left( 5.22 \right).......\left( i \right)$
For dealing the above trigonometric decimal, we use the approximate decimal expansion of $\pi $ as
$\Rightarrow \pi =3.14$
Multiplying both sides by $\dfrac{3}{2}$ we get
\[\begin{align}
  & \Rightarrow \dfrac{3\pi }{2}=\dfrac{3\left( 3.14 \right)}{2} \\
 & \Rightarrow \dfrac{3\pi }{2}=4.71.......\left( ii \right) \\
\end{align}\]
Adding and subtracting $4.71$ to the argument, $5.22$ we get
$\begin{align}
  & \Rightarrow 5.22=4.71+5.22-4.71 \\
 & \Rightarrow 5.22=4.71+0.51 \\
\end{align}$
Substituting the equation (ii) in the above equation, we grt
$\Rightarrow 5.22=\dfrac{3\pi }{2}+0.51........\left( iii \right)$
Now, we substitute the above equation (iii) in the equation (i) to get
$\Rightarrow E=\cos \left( \dfrac{3\pi }{2}+0.51 \right)$
We know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$. So the above expression can be written as
$\Rightarrow E=\cos \left( \dfrac{3\pi }{2} \right)\cos \left( 0.51 \right)-\sin \left( \dfrac{3\pi }{2} \right)\sin \left( 0.51 \right)$
Now since \[\sin \left( \dfrac{3\pi }{2} \right)=-1\] and \[\cos \left( \dfrac{3\pi }{2} \right)=0\] , we can write the above expression as
$\begin{align}
  & \Rightarrow E=\left( 0 \right)\cos \left( 0.51 \right)-\left( -1 \right)\sin \left( 0.51 \right) \\
 & \Rightarrow E=\sin \left( 0.51 \right).......\left( iv \right) \\
\end{align}$
Now, we know that the series expansion of the sine function is given by
$\Rightarrow \sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-......$
Substituting $x=0.51$ in the above expression we get
$\Rightarrow \sin \left( 0.51 \right)=\left( 0.51 \right)-\dfrac{{{\left( 0.51 \right)}^{3}}}{3!}+\dfrac{{{\left( 0.51 \right)}^{5}}}{5!}-......$
On solving the RHS, we get
$\Rightarrow \sin \left( 0.51 \right)=0.49$
Finally substituting the above in the equation (iv) we get
$\Rightarrow E=0.49$
Hence, we have found the value of the given trigonometric decimal as $0.49$.

Note:
We can also use the trigonometric identity $\cos \left( \dfrac{3\pi }{2}+x \right)=\sin x$ in the above solution. Also, while modifying the argument of the given expression, we must ensure that the remainder argument must be less than one. For example, we modified the argument in the above solution as $\dfrac{3\pi }{2}+0.51$ since $0.51$ is less than one.