Chapter 4

The relation Between the Trig Ratios of Some Specific Angles

The Trigonometric ratios of angle 2π + θ:

In the 1st Quadrant, all trig. ratios are positive.  Thinking of θ as an acute angle, θ by  itself ends in the 1st Quadrant at M.   2π + θ  ends in the 1st Quadrant at M as well, exactly at the same point on the unit circle where θ ends.  Note that means one full turn on the circle.  Both θ and 2π + θ will therefore have the same values of sine, cosine, tangent, and cotangent, as shown below:

 Important Formulas: sin(2π + θ) = sinθ, cos(2π + θ) = cosθ, tan(2π + θ) = tanθ, cot(2π + θ) = cotθ.

The Trigonometric ratios of angle 2π - θ:

Thinking of θ as an acute angle, ends in the 4th Quadrant where only cosine is positive.  means going 1 full turn from A to A and then returning to the M' that means as much as θ.  We may wirte:

 Another set of Important Formulas:   Thinking of θ as an acute angle, 2π -θ ends in the 4th Quadrant where only cosine is positive.  2π -θ means going 1 full turn from A to A and then returning to the M in the 4th Quadrant as much as θ.  We may wirte: sin(2π - θ) = -sinθ, cos(2π - θ) = +cosθ, tan(2π - θ) = -tanθ, cot(2π - θ) = -cotθ.

The Trigonometric ratios of angle π + θ:

 Thinking of θ as an acute angle (that ends in the 1st Quadrant), angle π + θ ends in the 3rd Quadrant where only tangent and cotangent are positive.  We may write: sin(π + θ) = -sinθ, cos(π + θ) = -cosθ, tan(π + θ) = +tanθ, cot(π + θ) = +cotθ. Connecting the M that is in the 3rd Quadrant to O and extending it to cross the tangent and cotangent axes, it crosses them in their positive region, and hence, both are positive for angle π + θ.

The Trigonometric ratios of angle π - θ:

 Thinking of θ as an acute angle (that ends in the 1st Quadrant), angle π - θ ends in the 2nd Quadrant where only sine is positive.  We may write: sin(π - θ) = +sinθ, cos(π - θ) = -cosθ, tan(π - θ) = -tanθ, cot(π - θ) = -cotθ.

The Trigonometric ratios of angle π/2 - θ:

Thinking of θ as an acute angle (that ends in the 1st Quadrant),/2 - θ) or (90°- θ) also ends in the 1st Quadrant.  Since in the 1st Quadrant, all trig. ratios are positive; therefore,  all trig ratios of  (π/2 - θ) angle are also positive.  What is the catch then?  Note that if two angles add up to 90°, they are called " complimentary angles.

If two angles add up to 90° or π/2, the sine of one is equal to the cosine of the other.

Also, the tangent of one is equal to the cotangent of the other.   θ and /2-θ) are complimentary angles because [ θ +  /2-θ) =  π/2 ].

By referring to the right triangle shown bellow, the proofs of the above statements are:

The Trigonometric ratios of angle π/2 + θ:

Thinking of θ as an acute angle (that ends in the 1st Quadrant),/2+θ) or (90°+θ) ends in the 2nd Quadrant where only sine of the angle is positive.    The /2+θ) formulas are similar to the /2-θ) formulas except only sine is positive because /2+θ) ends in the 2nd Quadrant.

sin/2+θ) = +cosθ.

cos/2+θ) = -sinθ.

tan/2+θ) = -cotθ.

cot/2+θ) = -tanθ.

Proof:

The Trigonometric ratios of angle (-θ):

Thinking of θ as an acute angle, (-θ) ends in the 4th Quadrant where only cosine is positive.  (-θ) means starting from A, the origin on the unit circle, and just going clockwise (negative direction) as much as θ.   Cosine being the only positive trig ratio in the 4th Quadrant, we may write:

 sin(-θ) = -sinθ cos(-θ) = +cosθ tan(-θ) = -tanθ cot(-θ) = -cotθ same as same as same as same as sin(2π - θ) = -sinθ. cos(2π - θ) = +cosθ. tan(2π - θ) = -tanθ. cot(2π - θ) = -cotθ. Note that (-θ) and  (2π-θ) end at the same point in the 4th Quadrant.

Example 1:

Verify the following equality:

cos(80°) -2sin(10°) + 3sin(190°) - 4cot(90°) + 5sin(350°) + 6sin(170°) = -3sin(10°).

Solution: Judging by 10° and 80°, we may conclude that all angles can be expressed in terms of the trig. ratios of 10°.  As usual, we will leave the right side untouched, and work on the left side to see if we get it equal to the right side.  We may write as

cos(90-10)° -2sin(10)° + 3sin(180+10)° - 4(0)° + 5sin(360-10)° + 6sin(180-10)° =

cos(π/2-10°) - 2sin(10°) + 3sin(π+10°) - 0 + 5sin(2π-10°) + 6sin(π-10°) =

+sin(10°) - 2sin(10°) + 3[-sin(10°)]  -0  +5[-sin(10°)] + 6[+sin(10°)] =

-3sin(10°)

This matches the right side. READ THE FOLLOWING CAREFULLY:

cos(80°) = ?  Ans.:   80° ends in 1st Quad. where cosine is (+); but cos(π/2 - 10°) = sin(10°) because 80° +10° = 90°  (complimentary angles)

-2sin(10°) = ?  Ans.:  We will leave this as it is because we re trying to convert all terms into sin(10°) if possible; thus, -2sin(10°) = -2sin(10°).

+3sin(190°) = ?  Ans.:  190° ends in the 3rd Quad. where sine is (-); thus,  3sin(π+10°) = -3sin10°.

-4cot(90°) = ?   Ans.:  -4(0) = 0.

5sin(350°) = ?   Ans.:  350° ends in the 4th Quad. where sine is (-); thus,  5sin(350°) = 5sin(2π-10°) = -5sin(10°).

6sin(170°) = ?  Ans.:  170° ends in the 2nd Quad. where sine is (+); thus,  6sin(170°) = 6sin(π-10°) = +6sin(10°).

Use the same style reasoning in the verification of the following exercises:

Exercises:

Without using a calculator, verify the following equalities: (Note: Work the right side only).

1) sin (200°) + 2sin(160°) - cos(70°) + 3sin(340°) - 4cos(110°) = sin(20°).

2) 3tan(130°) + cot(40°) - tan(230°) - 5tan(310°) -2cot(140°) = 4tan(50°).

3) 2cos(75°)cos(80°) + 3cos(105°)sin(170°) + 4sin(165°)sin(190°) =  -5sin(15°)sin(10°).

4) cos(115°) -3 sin(25°) - 5sin(205°) = cos(65°).

5) 3cos(310°) + cos(-50°) = 4cos(50°).

6) 2tan(60°) + 3tan(240°) = 5cot(30°).

7) sin(π + θ) + 2cos(π/2 - θ) -3sin(π - θ) = -2sin(θ).

8) 3tan(2π - θ) + tan(π + θ) + cot(π/2 - θ) - cot(π/2 + θ) = 0.