Trigonometry

Question

Given that  cosθ=-4/√53 and that the terminal side is in quadrant II, find the length of the third leg and  the other five trigonometric function values.

Answer 1

Remember that, in the coordinate plane, cosine is equivalent to a point's x-coordinate and sine is equivalent to a point's y-coordinate. So, in Quadrant II, sine is positive and cosine is negative. The other trigonometric functions are:

 

tan = sin/cos = +/- = negative

cot = cos/sin = -/+ = negative

sec = 1/cos = 1/- = negative

csc = 1/sin = 1/+ = positive

 

 

 

Now, let's find the third side of the triangle. Given that cos = adjacent/hypotenuse, we know that one leg has a length of 4 and the hypotenuse has a length of √(53). So:

 

a² + b² = c²

4² + b² = √(53)²

16 + b² = 53

b² = 37

b = √(37)

 

 

 

That's the value of the opposite side. Now that we know opposite = √(37), adjacent = 4, and hypotenuse = √(53), we can construct the other five functions:

 

sine (pos) = opposite/hypotenuse = √(37)/√(53)

tangent (neg) = -opposite/adjacent = -√(37)/4

cotangent (neg) = -adjacent/opposite = -4/√(37)

secant (neg) = -hypotenuse/adjacent = -√(53)/4

cosecant (pos) = hypotenuse/opposite = √(53)/√(37)

Answer 2

Cosine = adjacent/hypotenuse

 

We use the Pythagorean Theorem to find the third side.

 

(-4)² + x² = (√53)²

 

16 + x² = 53

 

x² = 37

 

x = √37

 

This is the length of the other leg.

 

sin θ = √37/√53

 

tan θ = -√37/4

 

csc θ = √53/√37

 

sec θ = -√53/4

 

cot θ = -4/√37