Find the greatest possible value of $5\cos x + 6\sin x$.
I attempted to solve this using graphing, however, the answer appears to be an ugly irrational. Is there a better method of solving this problem?
Thank you.
Find the greatest possible value of $5\cos x + 6\sin x$.
I attempted to solve this using graphing, however, the answer appears to be an ugly irrational. Is there a better method of solving this problem?
Thank you.
A simple vector-based approach: you recognize in the expression the dot product $$(5, 6) \cdot (\cos x, \sin x).$$ This product is maximized when the two vectors are parallel and it is then the product of the moduli $$\sqrt{5^2+6^2} \cdot 1.$$
Hint: Let $\theta$ be an angle whose sine is $\frac{5}{\sqrt{61}}$ and whose cosine is $\frac{6}{\sqrt{61}}$. Then our expression is equal to $$\sqrt{61}\sin(x+\theta).$$
There is an ad-hoc solution as shown in other answers.
Anyway, the standard approach is to find the roots of the derivative, $-5\sin x+6\cos x=0$, i.e. $\tan x=\frac{6}{5}$, so that $x=\arctan\frac{6}{5}$ or $\pi+\arctan\frac{6}{5}$.
Plug these values in the objective function.
Here's another way, by Cauchy-Schwarz inequality $$(\cos^2x+\sin^2x)(5^2+6^2)\ge (5\cos x + 6\sin x)^2$$
Let $r^2=5^2+6^2=25+36=61$ and $\alpha = \arctan \frac 56$
You will find that $$5\cos x+6\sin x=r(\sin\alpha\cos x+\cos\alpha \sin x)=r \sin (x+\alpha)$$