Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?


share|improve this question
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} $.

share|improve this answer
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.


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

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.