Question

Hey guys please help? Trigonometry

Answer 1

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

$\sin( \alpha ) = \frac{3 \sqrt{11} }{10}$

We must find cos.

Using the Pythagorean theorem

$( \sin( \alpha ) ) {}^{2} + ( \cos( \alpha ) ) {}^{2} = 1$

It is mostly notated as this,

$\sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1$

But they mean the same thing, we know

$\sin( \alpha ) = \frac{3 \sqrt{11} }{10}$

So we plug that in for sin a.

$( \frac{3 \sqrt{11} }{10} ) {}^{2} + \cos {}^{2} ( \alpha ) = 1$

$\frac{99}{100} + \cos {}^{2} ( \alpha ) = 1$

$\cos {}^{2} ( \alpha ) = \frac{100}{100} - \frac{99}{100}$

$\cos {}^{2} ( \alpha ) = \frac{1}{100}$

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

$\cos( \alpha ) = \frac{1}{10}$

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

$\cos( \alpha ) = - \frac{1}{10}$

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

$1 - \frac{99}{100}$

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

$\frac{100}{100} - \frac{99}{100} = \frac{1}{100}$