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Hi how do i go about solving this?

Find the values of the positive constants $k$ and $c$ such that $$-37\le k(3\sin\theta + 4\cos\theta) +c\le 43 $$for all values of $\theta$ $$\rightarrow-37\le k(5(\sin\theta + 53.1)) +c\le 43 $$ Then what?

Cheers

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Write $-37-c \le5k\sin(\phi)\le 43-c$. The range of the middle expression is $[-5k,5k]$. So if the inequality is "tight", you have $c=3$. Then solving for $k$ gives $k=8$. –  David Mitra Jan 22 '13 at 23:55

2 Answers 2

Hint: $|\sin \alpha| \leq 1$.

Hence, this implies that $ c = \frac {43+(-37)} {2} $.

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Hi cheers for the help! Where did the hint come from? Can you show me a step after? –  maxmitch Jan 22 '13 at 23:37
    
The hint is just stating the behavior of the $\sin$ function. The next step will be $ -5k \leq k 5 \sin(\theta + 53.1) \leq 5k$. –  Calvin Lin Jan 23 '13 at 0:34

By Cauchy Schwartz

$$(3\sin(x)+4\cos(x))^2 \leq 25 (\sin^2(x)+\cos^2(x))=25$$

and equality is possible.

Then

$$-5 \leq 3\sin(x)+4\cos(x) \leq 5 $$

this shows that

$$-5k+c \leq - k(3\sin\theta + 4\cos\theta) +c\le 5k+c \,.$$

and the lower/upper bounds can be atatined. You can finish it easely.

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