Chapter 5

The Relations Between the Sine, Cosine, Tangent, and Cotangent of an Angle

Important Trigonometric Identities:

We will attempt to derive a few important identities that relate the sine, cosine, tangent, and cotangent of an angle to each other.  Note that an identity holds true for all angles.  In doing so, we will take advantage of two formulas: (1) Pythagorean Theorem, and (2) the fact that the sum of angles in any triangle is 180°, that is A+B+C=180, especially for a right triangle that has A = 90°, and therefore B+C = 90°.  It is very important for you to know the derivations as follows: Identity (1):

Example 1:  Knowing that θ ends in the 2nd Quadrant and sinθ = 0.344,  find cosθ.

Solution: ( sinθ)2 + ( cosθ )2 = 1

(0.344)2 + ( cosθ )2 = 1

( cosθ )2 = 1 - 0.344)2 = 0.882

cosθ = ± 0.939 [The (-) answer is acceptable because in the 2nd Quad., cosine is (-)].

cosθ = - 0.939.

Example 2:  Knowing that β ends in the 3rd Quad., and cosβ = -3/5,  find sinβ.

Solution: ( sinβ)2 + ( cosβ )2 = 1

(sinβ)2 + ( -3/5 )2 = 1

( sinβ )2 = 1 - (-3/5)2 = 1 - 9/25 = 16/25

sinβ = ± 4/5 [The (-) answer is acceptable because in the 3rd Quad., sine is (-)].

sinβ = - 4/5.

Identity (2):

Example 3:  Given tanx = 0.25, find cotx.

Solution: since tangent and cotangent are the reciprocal of each other, we may write:

cotx = 1/tanx  = 1/0.25 = 4.0

Example 4:  Given cotC = 3/4, find tanC.

Solution: since tangent and cotangent are the reciprocal of each other, we may write:

tanC = 1/cotC  = 1/(3/4) = 4/3.

Identity (3):

Example 5:  Given sinD = 5/13, and cosD = 12/13, find tanD.

Solution: since sinD over cosD is equal to tanD, we may write:

tanD = sinD/cosD and therefore,  tanD = (5/13)/ (12/13) = 5/12.

Identity (4):

Note that since cosB/sinB = cotB, we mat square both sides to get:  cos2B/sin2B = cot2B  that is used in the following derivation:

Example 6:  Given sinD = 5/13, find cotD if D is an angle or arc that ends in the 1st Quadrant.

Solution: Using Identity (4), we have:

1 + cot2D = 1/(sin2D)    or,  1+ cot2D = 1/(25/169)    or,

cot2D = (169/25) - 1    or,

cot2D = 144/25    or,    cot2D = ±12/5   or,    cotD = -12/5.  (Why +?)

Note: Make sure that you solve the entire problem on your own but using horizontal fraction bars and not slashes.

Identity (5):

Example 7: Given tanQ = 3/4 and that Q ends in the 3rd Quadrant, find cosQ.

Solution:  Using the above identity, we have:

1 + tan2Q = 1/(cos2Q),    or,    1 + (3/4)2 = 1/(cos2Q).

25/16 = 1/(cos2Q),    or,    16/25 = cos2Q,    or    cosQ = ± 4/5

Since Q ends in the 3rd Quadrant where cosine is negative; therefore, cosQ = - 4/5.

================================================

Exercises:

Calculate the trig ratios of each of the following angles for which one trig ratio is given:

1) sin(m) = 5/7 and m ends in the 1st Quadrant.

2) sin(n) = -0.3 and n ends in the 3rd Quadrant.

3) cos(p) = (√8)/3 and p ends in the 4th Quadrant.

4) cos(q) = -8/11 and q ends in the 2nd Quadrant.

5) tan(θ) = √5 and θ ends in the 3rd Quadrant.

6) cot(h) = -2.4 and h ends in the 4th Quadrant.

7) sin(z) = (b-a)/(b+a),  and z ends in the 1st Quadrant.

To solve, write with horizontal fraction bar.

8) tan(u) = a/b and u ends in the 1st Quadrant.

To solve, first write with horizontal fraction bar.

9) Write the following expression in terms of sin(x), only:

5cos2x - 7tanx cosx

10) Write the following expression in terms of cos(x) only:

11) Write the following expression in terms of tan(x) only:

12) Calculate sin(x) once in terms of cos(x), and once in terms of tan(x).

13) Calculate cos(x) once in terms of sin(x), and once in terms of cot(x).

14) Calculate tan(x) once in terms of sin(x), and once in terms of cos(x).

15) Calculate the following fraction knowing that tan(x) = 5/12, and x ends in the 3rd Quadrant.

16) If sin(x) = 7/25, and x ends in the 2nd Quadrant, calculate the following:

z = (tanx + cotx)(cos2x - sin2x).

17) If sinx - cosx = 1/5, Calculate all trig ratios of angle x if  sinx - cosx = 1/5.

18) Find all trig ratios of angle x if x ends in the 1st Quadrant, and tanx - cotx = 7/12 .