Determine the constants a and b such that cos x-a sin x+b
Determine the constants $a$ and $b$ such that $\cos x-a\sin x+b\ln(1+x)-1=o(x^3)$ as $x\to0$.
Explanation of Method
Two unknowns can only silence two of the three lowest-order terms — not all three at once.
Step 1: Expand each function to order $x^3$
\[ \cos x = 1-\frac{x^2}2+o(x^3), \qquad \sin x = x-\frac{x^3}6+o(x^3), \qquad \ln(1+x) = x-\frac{x^2}2+\frac{x^3}3+o(x^3). \tag{1} \]
Step 2: Substitute and collect by power of $x$
\[ \cos x-a\sin x+b\ln(1+x)-1 = (-a+b)\,x + \left(-\frac12-\frac b2\right)x^2 + \left(\frac a6+\frac b3\right)x^3 + o(x^3). \tag{2} \]
Step 3: Try to kill all three coefficients
Vanishing of the $x^1$ and $x^2$ coefficients forces $b=a$ and then $b=-1,\,a=-1$ — a unique pair. But checking the $x^3$ coefficient at $a=b=-1$: \[\frac a6+\frac b3 = -\frac16-\frac13 = -\frac12 \neq 0. \tag{3}\]
Two free parameters give two equations to satisfy, but there are three coefficients ($x^1,x^2,x^3$) to silence; generically this is impossible, and indeed it fails here.
Step 4: Interpretation
The values $a=b=-1$ are exactly the ones that silence the two lowest orders, making the expression as small as possible near $x=0$: $o(x^2)$, of exact order $x^3$ (not the stronger $o(x^3)$ literally asked for, which no choice of $a,b$ can achieve).
Step 5: Verification
\[x=0.05\text{: }\cos x+\sin x-\ln(1+x)-1\approx-0.0000607,\ -x^3/2=-0.0000625\ \checkmark\]
Result
Final answer:
\[\boxed{\,a=-1, \qquad b=-1\,}\]
With these values: $\cos x+\sin x-\ln(1+x)-1=-\dfrac{x^3}2+o(x^3)$ — the best attainable order with two free parameters.
Remark
Remark. The problem's own phrasing $o(x^3)$ is, strictly, unsatisfiable by any pair $(a,b)$: with two free parameters one can silence exactly two leading coefficients, landing on $O(x^3)$ (not $o(x^3)$) at best. This is not a computational slip in the source but an inherent limit of the setup; the intended reading is "make the expression vanish as fast as possible," which is exactly $a=b=-1$.