Hey guys please help? Trigonometry So, the task is: find cos a, if:

1) sin a = (3√11)/10, a ∈ (0; π/2)

2) sin a = (3√11)/10, a ∈ (π/2; π)

Could someone please explain how to solve it? I can't figure out what difference a ∈ (0; π/2) and a ∈ (π/2; π) make in the way I have to solve it mmh... I'll pin my attempt to do the first one (failed for some reason)

Step-by-step explanation:

a ∈ (0; π/2) here means that our angle, a must lie between 0 and pi/2, exclusive.

So this mean our angle must be in between 0 and pi/2, but can not be neither 0 and pi/2.

Here we have

We must find cos.

Using the Pythagorean theorem

It is mostly notated as this,

But they mean the same thing, we know

So we plug that in for sin a.

Since cos is Positve over the interval (0; π/2), we take the positive or principal square root.

2. We would get the same work for the second part, the only difference is that cosine is negative over the interval

(π/2, π)

So the answer for 2 is

Disclaimer: Your work you did was correct, just remember for fractions like

Convert 1 into a fraction that has a denominator of 100.

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints