Evaluating Trigonometric Functions

[nycmathguy]

Section 5.1

Questions 10

cos (theta) = 2/3, sin (theta) < 0

I see that cos (theta) is positive and sin (theta) is negative. This tells me that we are in quadrant 2.

Now, cosine is adjacent over hypotenuse.

Let x = opposite side of a right triangle in quadrant 2.

x^2 + 2^2 = 3^2

x^2 = 9 - 4

x^2 = 5

sqrt{x^2} = sqrt{5}

x = sqrt{5}

I can now find the remaining 6 trigonometric functions of theta.

NOTE: I WILL NOT RATIONALIZE THE DENOMINATOR.

sin (theta) = -sqrt{5}/3

cos (theta) = 2/3 as given.

tan (theta) = sqrt{5}/2

csc (theta) = -(3/sqrt{5})

sec (theta) = 3/2

cot (theta) = 2/sqrt{5}

You say?

 

[MathLover1]

correct

 

[nycmathguy]

correct

No need to answer another one just like it. Moving on....