# 1 The sum of the three smallest positive values of theta such that 4(cos*theta)(sin*theta) = 1 is k*pi. Find k?

May 16, 2018

$4 \cos \theta \sin \theta = 1$

$\implies 2 \cos \theta \sin \theta = \frac{1}{2}$

$\implies \sin 2 \theta = \sin \left(\frac{\pi}{6}\right)$

$\implies 2 \theta = n \pi + {\left(- 1\right)}^{n} \cdot \left(\frac{\pi}{6}\right)$

$\implies \theta = \frac{1}{2} \left(n \pi + {\left(- 1\right)}^{n} \cdot \left(\frac{\pi}{6}\right)\right) \text{where } n \in \mathbb{Z}$

When $n = 0 \to \theta = \frac{\pi}{12}$

When $n = 1 \to \theta = \frac{5 \pi}{12}$

When $n = 2 \to \theta = \frac{13 \pi}{12}$

By the condition of the problem

$\frac{\pi}{12} + \frac{5 \pi}{12} + \frac{13 \pi}{12} = k \pi$

So $k = \frac{19}{12}$

May 16, 2018

$k = \frac{19}{12}$

#### Explanation:

We can rewrite as

$2 \cos \theta \sin \theta = \frac{1}{2}$

$\sin \left(2 \theta\right) = \frac{1}{2}$

Now I'm assuming that we're only consider values of $\theta$ greater than $0$.

$2 \theta = \frac{\pi}{6} , \frac{5 \pi}{6} , \frac{13 \pi}{6}$

$\theta = \frac{\pi}{12} , \frac{5 \pi}{12} , \frac{13 \pi}{12}$

Now you just have to add the values up

${\theta}_{\text{sum}} = \frac{19 \pi}{12}$

Therefore $k = \frac{19}{12}$

Hopefully this helps!

May 16, 2018

$k = \frac{19}{12}$

#### Explanation:

Here,

$4 \cos \theta \sin \theta = 1$

$\implies 2 \sin \theta \cos \theta = \frac{1}{2}$

$\implies \sin 2 \theta = = \frac{1}{2} > 0 \implies {I}^{s t} Q u a \mathrm{dr} a n t \mathmr{and} I {I}^{n d} Q u a \mathrm{dr} a n t$

$\implies 2 \theta = \frac{\pi}{6} , \left(\pi - \frac{\pi}{6}\right) , \left(2 \pi + \frac{\pi}{6}\right) , \left(3 \pi - \frac{\pi}{6}\right) , \left(4 \pi + \frac{\pi}{6}\right) , \ldots$

$\implies 2 \theta = \frac{\pi}{6} , \frac{5 \pi}{6} , \frac{13 \pi}{6} , \frac{17 \pi}{6} , \frac{25 \pi}{6} , \frac{29 \pi}{6} , \ldots$

$\implies \theta = \frac{\pi}{12} , \frac{5 \pi}{12} , \frac{13 \pi}{12} , \frac{17 \pi}{12} , \ldots$

The sum of the three smallest positive values of theta $= k \pi$

So,

$\frac{\pi}{12} + \frac{5 \pi}{12} + \frac{13 \pi}{12} = k \pi$

$\implies \frac{\pi + 5 \pi + 13 \pi}{12} = k \pi$

$\implies \frac{19 \pi}{12} = k \pi$

$\implies \frac{19}{12} = k$

$i . e . k = \frac{19}{12} \approx 1.5833$