Question

Let (6,-5) be a point on the terminal side of θ. Find the exact values of sin θ, sec θ, and tan θ.

Answer 1

Answer:

sinθ = -5/√61

secθ = √61/6

tanθ = -5/6

Step-by-step explanation:

From the given coordinate (6, -5), x = 6 and y = -5. This shows that the point lies in the 4th quadrant. In the fourth quadrant, only cos θ is positive, both sin θ and tan θ are negative.

Let us get the value of the radius 'r' first before calculating the trigonometry identities.

Using the Pythagoras theorem;

$r^2 = x^2 + y^2\\r^2 = 6^2 + (-5)^2\\r^2 = 36+25\\r^2 = 61\\r = \sqrt{61}$

Using SOH, CAH, TOA to get the trigonometry identities;

Given x = 6, y =5 and r = √61

$sin \theta = opp/hyp = y/r\\sin \theta = 5/\sqrt{61}$

Since sin θ, is negative in the fourth quadrant, $sin \theta = -5/\sqrt{61}$

$cos \theta = adj/hyp = x/r\\cos \theta = 6/\sqrt{61}$

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

For tanθ:

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

Since tan θ, is negative in the fourth quadrant, $tan \theta = -5/6$