Question

question 1 and 2 please!!

2. The terminal side of an angle o in standard position lies on the line in the given quadrant. Find the sine and cosine functions of 0. y=-3x Quadrant IV sin 0-(Type an exact answer, using radicals as needed.) 3. The terminal side of an angle o in standard position lies on the line in the given quadrant. Find the sine and cosine functions of 0. y = 3x Quadrant III sin 0=(Type an exact answer, using radicals as needed.)

Answer 1

Solution-

(a)

Consider the equation of line y = -3x

If a point lies along the terminal side of this line in IVth quadrant(as given in the question) than

Let a point

(1, -3) lies in IVth quadrant along the given line.

Now, the value of sin and cosine will be same along the given as with the point.

So,

Y sin(O)= -3 (1)2 + (-3) -3 1 +9 3 V10 2 + y2

and$cos(\Theta )=\frac{x}{\sqrt{x^2+y^2}}=\frac{1}{\sqrt{(1)^2+(-3)^2}}=\frac{1}{1+9}=\frac{1}{\sqrt{10}}$

Hence,

,

3 sin(O) = /10

$cos(\Theta )=\frac{1}{\sqrt{10}}$

(b)

Consider the equation of line y = 3x

If a point lies along the terminal side of this line in IIIrd quadrant(as given in the question) than

Let a point

(-1, -3) lies in IIIrd quadrant along the given line.

Now, the value of sin and cosine will be same along the given as with the point.

So,

$sin(\Theta )=\frac{y}{\sqrt{x^2+y^2}}=\frac{-3}{\sqrt{(-1)^2+(-3)^2}}=\frac{-3}{1+9}=-\frac{3}{\sqrt{10}}$

and

$cos(\Theta )=\frac{x}{\sqrt{x^2+y^2}}=\frac{-1}{\sqrt{(-1)^2+(-3)^2}}=\frac{-1}{1+9}=-\frac{1}{\sqrt{10}}$

Hence,

3 sin(O) = /10
,

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