knowing: $\cos x+\sin x=\frac{5}{4}$, obtain: $\cos(4x)$ 
knowing: $\cos x+\sin x=\frac{5}{4}$,
obtain: $\cos(4x)$

$$\cos x+\sin x=\frac{5}{4}$$
$$\sin^2x+\cos^2x+2\sin x\cos x=\frac{25}{16}$$
$$\sin2x=\frac{25}{16}-\frac{16}{16}=\frac{9}{16}$$
$$\cos4x=1-2\sin^22x=1-2\Bigl(\frac{9}{16}\Bigr)^2=\frac{47}{128}$$
Taken out of one of the TAU entry tests but unfortunately, they don't give solutions to most of the exercises, so… Am I correct?
 A: Your computation is correct, and rather elegant.
A point that may not go down so well is that your argument consists only of four equations, without verbal explanations or an indication of the logical connections between these equations. However, the argument is short and transparent, I don't think it would be made more readable or easier to follow if wrapped in words like "By squaring both sides of the premise $$\cos x + \sin x = \frac{5}{4}$$ we obtain the equation $$\cos^2 x + \sin^2 x + 2\sin x \cos x = \frac{25}{16}$$ which, after using the trigonometric identity $\sin^2 x + \cos^2 x = 1$ and rearranging yields …".
On the contrary, in this case that would only add clutter, in my opinion. But prefacing it with a remark along the lines of "In the following, every equation follows from the previous by applying an elementary operation and/or trigonometric identity" would not be detrimental.
A: Another way to do this is as follows-
$\cos x+\sin x=\frac{5}{4}$, obtain: $\cos(4x)$
$$\Rightarrow 4\cos x + 4\sin x = 5$$
$$\Rightarrow 16 \cos^2x +25-40 \cos x = 16 \sin x$$
$$\Rightarrow 15 \cos^2x -56 \sin x+25=0$$
$$\Rightarrow \cos x = \frac{\sqrt{409}}{15}$$
Now, by knowing the value of x, the cosine value of $4x$ can be calculated in the similar manner.
