Respuesta :
The functions are illustrations of trigonometry ratios
[tex]\mathbf{(a)\ cos\theta = \frac{\sqrt{15}}{5}}[/tex]
The sine of an angle is calculated as follows
[tex]\mathbf{sin\theta = \sqrt{1 - cos^2\theta}}[/tex]
So, we have:
[tex]\mathbf{sin\theta = \sqrt{1 - (\sqrt{15}/5)}}[/tex]
[tex]\mathbf{sin\theta = \sqrt{1 - (15/25)}}[/tex]
Take LCM
[tex]\mathbf{sin\theta = \sqrt{\frac{10}{25}}}[/tex]
[tex]\mathbf{sin\theta = \frac{\sqrt{10}}{5}}[/tex]
[tex]\mathbf{(b)\ tan\theta = \frac{\sqrt{10}}{2}}[/tex]
The tangent of an angle is:
[tex]\mathbf{\ tan\theta = \frac{Opposite}{Adjacent}}[/tex]
Using Pythagoras theorem, we have:
[tex]\mathbf{Hypotenuse^2 = Opposite^2 + Adjacent^2}[/tex]
So, we have:
[tex]\mathbf{Hypotenuse^2 = (\sqrt{10})^2 + 2^2}[/tex]
[tex]\mathbf{Hypotenuse^2 = 10 + 4}[/tex]
[tex]\mathbf{Hypotenuse^2 = 14}[/tex]
Take square roots
[tex]\mathbf{Hypotenuse = \sqrt{14}}[/tex]
So, the sine of the angle is:
[tex]\mathbf{\ sin\theta = \frac{Opposite}{Hypotenuse}}[/tex]
[tex]\mathbf{\ sin\theta = \frac{\sqrt{10}}{\sqrt{14}}}[/tex]
Rationalize
[tex]\mathbf{\ sin\theta = \frac{\sqrt{140}}{14}}[/tex]
[tex]\mathbf{(c)\ csc\theta = \frac{\sqrt{15}}{3}}[/tex]
The sine of the angles is:
[tex]\mathbf{sin \theta = \frac{1}{csc\theta}}[/tex]
So, we have:
[tex]\mathbf{sin \theta = \frac{1}{\sqrt{15}/3}}[/tex]
[tex]\mathbf{sin \theta = \frac{3}{\sqrt{15}}}[/tex]
Rationalize
[tex]\mathbf{sin \theta = \frac{3\sqrt{15}}{15}}[/tex]
[tex]\mathbf{sin \theta = \frac{\sqrt{15}}{5}}[/tex]
[tex]\mathbf{(d)\ sec\theta =\frac{\sqrt 6}{3}}}[/tex]
The sine of an angle is calculated as follows
[tex]\mathbf{sin\theta = \sqrt{1 - (1/sec\theta)^2}}[/tex]
So, we have:
[tex]\mathbf{sin\theta = \sqrt{1 - (3/\sqrt 6)^2}}[/tex]
[tex]\mathbf{sin\theta = \sqrt{(1 - 9/6)}}[/tex]
[tex]\mathbf{sin\theta = \sqrt{( - 3/6)}}[/tex]
The sine of the angle does not have a real value
[tex]\mathbf{(e)\ cot\theta = \frac{\sqrt 6}{2}}[/tex]
Start by calculating the tangent of the angle
[tex]\mathbf{tan\theta = \frac{1}{cot\theta}}[/tex]
So, we have:
[tex]\mathbf{tan\theta = \frac{1}{\sqrt 6/2}}[/tex]
[tex]\mathbf{tan\theta = \frac{2}{\sqrt 6}}[/tex]
Rationalize
[tex]\mathbf{tan\theta = \frac{2\sqrt 6}{6}}[/tex]
[tex]\mathbf{tan\theta = \frac{\sqrt 6}{3}}[/tex]
The tangent of an angle is:
[tex]\mathbf{\ tan\theta = \frac{Opposite}{Adjacent}}[/tex]
Using Pythagoras theorem, we have:
[tex]\mathbf{Hypotenuse^2 = Opposite^2 + Adjacent^2}[/tex]
So, we have:
[tex]\mathbf{Hypotenuse^2 = (\sqrt{6})^2 + 3^2}[/tex]
[tex]\mathbf{Hypotenuse^2 = 6 + 9}[/tex]
[tex]\mathbf{Hypotenuse^2 = 15}[/tex]
Take square roots
[tex]\mathbf{Hypotenuse = \sqrt{15}}[/tex]
So, the sine of the angle is:
[tex]\mathbf{\ sin\theta = \frac{Opposite}{Hypotenuse}}[/tex]
[tex]\mathbf{\ sin\theta = \frac{\sqrt{6}}{\sqrt{15}}}[/tex]
Rationalize
[tex]\mathbf{\ sin\theta = \frac{\sqrt{90}}{15}}[/tex]
Rewrite as:
[tex]\mathbf{\ sin\theta = \frac{3\sqrt{10}}{15}}[/tex]
[tex]\mathbf{\ sin\theta = \frac{\sqrt{10}}{5}}[/tex]
Read more about trigonometry ratios at:
https://brainly.com/question/24888715
