<s>[INST] 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?
 [/INST] 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.
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