Chapter 3

Variations of Sine, Cosine, Tangent, and Cotangent of an Angle

1) Variations of Sine:

 It is easy to see how the sine of an angle ( sinθ ) varies as angle θ varies from 0 to 2π (One full cycle).  In the unit circle shown, let angle θ or Arc AM vary from 0 to 360° or 2π.  At any position that M has on the circle, draw a line perpendicular to the x-axis and call it MP, as shown.  It is easy to see that MP measures the sine of angle θ or the sine of Arc AM.  We may therefore write: Sinθ = (opp./hyp.) = (MP/1) = MP. (MP or sinθ is a ratio and therefore dimensionless).

 Examine the following 12 figures in the order shown, and in any of the cases verify that

Conclusion:

The chart of variations of sinθ vesus θ from 0 to 2π as well as the corresponding graph are shown below:

Example 1: Verify the following equality:

sin(90) +2sin(0°) + 3sin(180°) -4sin(270) = 5

Solution:  Always, leave one side untouched, and work on the other side until both sides prove to be equal.  In this example, the left side can be worked on; therefore, we will leave the right side alone, and work on the left side as follows:

1 + 2(0) + 3(0) -4(-1) = 5, or

5 = 5. (Verified).

Exercises: Verify the following equalities:

1) 3sin(270) - 4sin(90) + 250sin(180) - 345sin(360) = -7.

2) -6sin(π) + 5sin(2π) -8sin(3π/2) + 5sin(π/2) = 13.

3) -23sin(3π/2)  - 20sin(π/2) + 4sin(2π) = 3.

2) Variations of Cosine:

 Now let's study the variations of the cosine of an angle ( cosθ ) as angle θ varies from 0 to 2π.  In the unit circle shown, let angle θ or Arc AM vary from 0 to 360° or 2π.  At any position that M has on the circle, draw a line perpendicular to the x-axis and call it MP, as shown.  It is easy to see that OP measures the cosine of angle θ or Arc AM.  We may therefore write: cosθ = (adj./hyp.) = (OP/1) = OP. (OP or  cosθ is a ratio and therefore dimensionless).

Track the variations of cosθ or OP in the following figures:

 Examine the following 12 figures in the order shown, and in any of the cases verify that

Conclusion:

The chart of variations of cosθ vesus θ from 0 to 2π as well as the corresponding graph are shown below:

Example 2: Verify the following equality:

-cos(90) -5cos(0°) + 6cos(180°) + 45cos(270) = -11.

Solution:  Always, leave one side untouched, and work on the other side until both sides prove to be equal.  In this example, the left side can be worked on; therefore, we will leave the right side alone, and work on the left side as follows:

-0 - 5(1) + 6(-1) + 45(0) = -11, or

-11 = -11. (Verified).

Exercises: Verify the following equalities:

4) 5cos(270) -7cos(90) + 50cos(180) -45cos(360) + 4cos(0) = -91

5) -6cos(π) + 5cos(2π) -8cos(3π/2) + 5cos(π/2) = 11.

6) -23cos(3π/2)  + 20cos(π/2) + 25cos(2π)  - cos(0) = 24.

3) Variations of Tangent:

 From Point A, the origin of the Trig. Circle, draw line u'Au tangent to the circle, as shown.  u'Au  may be called the "Axis of Tangents."  The reason will become apparent shortly.  For a typical angle θ, as shown, let's extend OM until it crosses the u'Au axis at T.   In triangle OAT, we may write: tanθ = Opp./Adj. = AT / 1 = AT.  (Note that AT or  tanθ  is a ratio and thus dimensionless). As θ increases while arc AM grows and M moves counterclockwise on the circle, Point T moves on the u'Au axis.   For different values of angle θ, as shown, we will have different values of AT or tanθ.  Note that as θ approaches 90°, point T moves up the Au axis faster and faster and AT or tanθ becomes greater and greater in value.  At exactly θ = 90°, OM becomes parallel to the Au axis and does not cross it; therefore, tan90° does not exist and is undefined.  For angles less than 90° but extremely close to 90° ( shown as 90-), the two lines cross very far up and AT or  tanθ becomes infinitely large, say +∞.  We may write: tan 90- = +∞   (Better to say goes to +∞, or becomes unbound.) For angles extremely close to 90 but greater than 90 (shown as 90+), OM does not cross the Au part.  We have to extend OM in the opposite direction to cross the u'A part of the tangents axis at points very far down or -∞.  While M is in the 2nd Quadrant, extending OM in the opposite direction crosses the tangents axis in its negative part (the u'A part) and tangent is negative.  The conclusion is that: If an arc or angle ends in the 2nd Quadrant, its tangent is negative.  The following figures show the variations of tanθ as θ changes from 0 to 2π. If an arc barely passes 90°, its tangent is a very big negative number (-∞). tan 90+ = -∞   (Better to say goes to -∞ or becomes unbound but negative.)

In the following 12 figures, the variations of tanθ are shown, 3 figures per quadrant:

 In the 1st Quadrant, as  θ varies from 0 to 90°, tanθ or AT varies from 0 to ∞,and is positive. When M is at B, θ = 90°; OM is parallel to u'Au; tan90° = undef. In the 2nd Quadrant, just past 90°, tangent value jumps to -∞. tan90+ = -∞.   While 90°<θ<180°, tanθ is (-)   As θ nears 180°, tanθ goes to 0, and tan180° = 0. In the 3rd Quadrant, where θ varies from 180° to 270°, tanθ is (+), and acts as if θ varies from 0° to 90°. In the 4th Quadrant, When θ is between 270° and 360°, OM crosses the u'A axis making AT or tanθ  negative.   As θ approaches 360°, AT or tanθ goes to 0. tan 360° = 0.

The variations of  tanθ as θ varies from 0 to as well as its graph are show below:

Example 3: Verify the following equality:

-4tan(0) -2tan(180°) - 9cos(180°) + 3sin(270) = 6.

Solution:  Always, leave one side untouched, and work on the other side until both sides prove to be equal.  In this example, the left side can be worked on; therefore, we will leave the right side alone, and work on the left side as follows:

-4(0) - 2(0) - 9(-1) + 3(-1) = 6, or

+ 6 = + 6. (Verified).

Exercises: Verify the following equalities:

7)   5sin(270) -8cos(90) + 50tan(180) -45tan(360) + 4tan(0) = -13.

8)  -4tan(π) - 3tan(2π) -8cos(3π/2) + 5sin(π/2) = 5.

9)  -2tan(2π)  + 17sin(π/2) + 25tan(π)  - tan(0) = 17.

4) Variations of Cotangent:

rom Point B, draw line v'Bv tangent to the circle, as shown.  v'Bv  may be called the "Axis of Cotangents."  The reason will become apparent shortly.  For a typical angle θ, as shown, let's extend OM until it crosses the v'Bv axis at K  In triangle OBK, we may write:  cotθ = BK/1 = BK.

(Note that BK or  cotθ  is a ratio and thus dimensionless).

As θ increases while arc AM grows (as M moves on the circle counterclockwise ), Point K moves on the v'Bv axis.   For different values of angle θ, as shown, there are different values of BK or cotθ.   If M is moved back to A, θ → 0, and OM becomes parallel to the the v'Bv axis and does not cross it while K has to go so far and suddenly disappear; therefore,

cot( 0°) = undefined.

For angles extremely close , but positive( 0+ ), Point K is extremely far from B to the right; therefore,

cot( 0+) = +  (Better to write as θ → 0+, BK or cotθ → +).

As θ increases from 0 to 90° in the 1st Quadrant, Point K moves from + to B that means cotθ decreases from + to 0.  We may write:

cot(90°) = 0

When M enters the 2nd Quadrant, Point K moves in the negative portion of the axis of cotangents (v'B).

If M barely reaches 180°, its cotangent becomes a very big negative number (-∞).

tan 180- = -∞   (Better to write as θ → 180-, BK or cotθ → -∞).

The following 12 figures show the variations of cotθ as θ changes from 0 to 2π:

 In the 1st Quadrant, as  θ varies from 0 to 90°, cotθ or BK varies from +∞ to 0,and cotangent is positive. When M is at B, θ = 90°; BK = 0; therefore, cot (90°) = 0. In the 2nd Quadrant, as θ exceeds 90°,  BK or cotθ becomes negative. When θ→ 180-, cot (180-) → -∞. If  θ = 180° exactly, cot (180°) = undef. In the 3rd Quadrant, when arc AM passes point C, extending OM does not cross the v'Bv axis and it has to be extended in the opposite direction to cross the v'Bv axis in the 1st Quadrant; thus, when θ varies from 180° to 270°, cotθ is (+), and acts as if θ varies from 0° to 90°. In the 4th Quadrant, When θ is between 270° and 360°, again, OM has to be extended in the opposite direction to cross the cotangents axis. The extension in opposite direction crosses v'Bv in its (-) portion; therefore, in the 4th Quadrant, BK or cotθ  is negative. As θ approaches 360°, BK or cotθ goes to -∞. cot (360-) = -∞, finally,cot (360) = cot (0) = undef.

The variations of  cotθ as θ varies from 0 to as well as its graph are show below:

Example 4: Verify the following equality:

3cot(90) -3sin(90°) - 4cos(180°) -9sin(270) = 10

Solution:  Always, leave one side untouched, and work on the other side until both sides prove to be equal.  In this example, the left side can be worked on; therefore, we will leave the right side alone, and work on the left side as follows:

3(0) - 3(1) - 4(-1) - 9(-1) = 10, or

10 = 10. (Verified).

Exercises: Verify the following equalities:

10) 2cot(270) -4cot(90) + 250cos(180) -450sin(270) = +200.

11) -6tan(π) + 2cos(2π) -4cot(3π/2) + 5sin(π/2) + tan(0) = 7.

12) -33cot(3π/2)  - 20cot(π/2) + 4tan(2π) - 3tan(0) = 0.

9) Summary:

From the above 4 analyses, we may conclude that:

 1) If θ ends in the 1st quadrant, sinθ, cosθ, tanθ, cotθ  are all (+). 2) If θ ends in the 2nd quadrant, only sinθ  is (+). 3) If θ ends in the 3rd quadrant, tanθ and cotθ are (+). 4) If θ ends in the 4th quadrant, only cosθ  is (+). All other cases are (-)