Question

Find the value of the six trigonometric functions of 0, where 0 angle formed by positive x-axis and line segment from (0,0)to(6,5).

Sin(0)=_________
cos(0)=_________
tan(0)=_________
sec(0)=_________
csc(0)=_________
cot(0)=_________

Answer 1

Hello!

It's important to use a proper symbol for the argument of the trig functions.  When you typed "0," you were possibly thinking of the Greek letter theta, ∅.  I'll assume that you meant "theta" here.

The line segment connecting (0,0) and (6,5) is the hypotenuse of a right triangle; its length, found by applying the Pythagorean Theorem, is sqrt( (6-0)^2 + (5 -0)^2 ), or sqrt(36+25), or sqrt(61).   Opposite angle ∅ is the "opp side," connecting (6,5) to the x-axis; its length is 5.  Finally, the horizontal line segment connecting (0,0) and (6,0) is the "adj" side;" its length is 6.

Thus:  sin ∅ = opp / hyp = 5/√61

cos ∅ = adj / hyp = 6/√61

tan ∅ = opp / adj = 5/6

cot ∅ = 1/tan ∅ = 6/5

sec ∅ = hyp / adj = √61 / 6 (same as 1/cos ∅)

csc ∅ = hyp / opp = √61 / 5 (same as 1/sin ∅)

Answer 2

The values of the 6 trigonometric ratios of $\theta$ are:

$sin\theta=\frac{5}{\sqrt{61}}$

$cos\theta=\frac{6}{\sqrt{61}}$

$tan\theta=\frac{5}{6}}$

$sec\theta=\frac{\sqrt{61}}{6}$

$csc\theta=\frac{\sqrt{61}}{5}$

$cot\theta=\frac{6}{5}$

First, we need to find the distance from the origin to the point $(x,y)=(6, 5)$. That distance, $r$, is the hypotenuse.

$r=\sqrt{x^2+y^2}\\=\sqrt{6^2+5^2}\\=\sqrt{61}$

Next, we need to compute the values of the 6 trigonometric ratios of $\theta$:

$sin\theta=\frac{y}{r}=\frac{5}{\sqrt{61}}$

$cos\theta=\frac{x}{r}=\frac{6}{\sqrt{61}}$

$tan\theta=\frac{y}{x}=\frac{5}{6}}$

$sec\theta=\frac{1}{cos\theta}=\frac{\sqrt{61}}{6}$

$csc\theta=\frac{1}{sin\theta}=\frac{\sqrt{61}}{5}$

$cot\theta=\frac{1}{tan\theta}=\frac{6}{5}$

Learn more about trigonometric ratios here: https://brainly.com/question/1201366