# What is maximum value of $f(t)=16\cos t \cdot \cos 2t \cdot\cos 3t \cdot \cos6t$

What is maximum value of $$f(t)=16\cos t \cdot \cos 2t \cdot\cos 3t \cdot \cos6t$$

My Approach:

$$1. \;\;$$Directly $$t=0$$ gives me maximum value of $$16$$.

$$2.\;\;$$ Converted it into $$f(t)=\dfrac{\sin(4t)\sin(12t)}{\sin (t)\sin(3t)}\;\;$$ but couldn't proceed further from this step.

My Doubts: $$1.\;$$Can we get maximum value without putting $$t=0\;?$$

$$2.\;$$ How can i proceed further using second method to get maximum value?

$$3.\;$$ Is there other way to solve this problem ?

• I don't understand why you don't want to put $t=0$. Trying other methods makes no sense. Commented May 30, 2022 at 5:28
• For your first question: since $\cos$ has period $2 \pi$, your function $f$ also has period $2 \pi$, so the answer is "yes": any integer multiple of $2 \pi$ will also yield 16. Dunno about your other two questions, though. Commented May 30, 2022 at 5:37
• Don't know why i am getting downvotes to question. Commented May 30, 2022 at 6:25

Observe that $$\forall t, f(t) \le 16$$ and $$f(t) = 16$$ when $$t = 0$$. This shows $$16$$ is the max value of $$f$$. All you need is $$\cos(kt) \le 1$$ for any $$k,t$$.

• Two downvotes..? Just why, I don't think there is any error in this answer and this answers the question as well. Commented May 30, 2022 at 5:54
• The answer is perfectly fine! Commented May 30, 2022 at 6:14
• also don't know why downvotes to question too. Commented May 30, 2022 at 6:26

I think you did not answer the interesting part of the problem: why is $$t=0$$ a maximum of $$f$$? Wang YeFei did answer this.

1. You cannot get the maximum value without evaluating the function at a maximum.
2. Your conversion of $$f$$ is not even defined at $$t=0$$. So this way can not work.
3. The usual way would be to find $$f'$$ to get all local extrema and pick the one with highest value as global maximum. But this would be unnecessaryly complicated here…

1. Yes, but it would involve setting $$f'(t)=0$$ - you can use logarithmic differentiation$$^1$$ to get the derivative or use the product rule multiple times.
$$^1$$ To find the logarithmic derivative, we would have...
$$f(t) = 16 \cos t \cos 2t \cos 3t \cos 6t \\ \ln f(t) = \ln 16 + \ln \cos t + \ln \cos 2t + \ln \cos 3t + \ln \cos 6t \\ \frac {f'(t)}{f(t)} = 0 + \frac {1}{\cos t}(-\sin t) + \frac {1}{\cos 2t}(-2\sin 2t) + \frac {1}{\cos 3t}(-3\sin 3t) + \frac {1}{\cos 6t}(-6\sin t) \\ f'(t) = f(t) (-\tan t -2 \tan 2t - 3 \tan 3t - 6 \tan 6t) \\ f'(t) = 16 \cos t \cos 2t \cos 3t \cos 6t(-(\tan t +2 \tan 2t + 3 \tan 3t + 6 \tan 6t))$$ Setting $$f'(t) = 0$$, you would have either $$16 \cos t \cos 2t \cos 3t \cos 6t = 0$$ or $$-(\tan t +2 \tan 2t + 3 \tan 3t + 6 \tan 6t)= 0$$