We know that both sin and cos are seen in the ratio identities of the other 4 trig functions. Before examining their relations to each other, we need

**Tangent?**

The ratio for tangent is sin/cos aka y/x. So the asymptotes are located when tangent is cosine, this would be where cos is equal to 0. Because cos is the x-value, x=0 at pi/2 and 3pi/2this is where the asymptotes are. Knowing this, we will be looking at the relations between the graphs of tan, cos, and sin. Let's begin with the first quadrant. We know that on the unit circle, Q1 begins at 0 and ends at pi/2. According to ASTC, we know that all trig functions in the first quadrant are positive. However, let's pretend that we don't know that tangent is positive in the first quadrant, how can we find out where tangent belongs? We can use the trig ratio for tan: sin/cos. Both sin and cos are pos. in the first quadrant, when you divide one by the other, you end up with a pos. value for tangent. For Q2, from pi/2 to pi, we know that sin is pos. and cos is neg. This means tangent is neg. in the second Qaudrant because pos/neg=neg. Q3 is from pi to 3pi/2, in which sin and cos are both neg. A neg/neg gives us a pos. value for tan. In Q4, from 3pi/2 to pi, sin is pos and cos is neg. A pos/neg gives us a neg. value for tangent in the 4th quadrant. Thus, the graph for tangent is pos, neg, pos, neg. When we graph it within its period of pi/2 to 3pi/2, the graph will be uphill. In its 1st period, it will start below the x-axis close to the asymptote at pi/2 and move uphill when it enters into Q3 where it is positive. This pattern will continue for every tangent graph.

https://www.desmos.com/calculator/hjts26gwst |

Cotangent?

The trig ratio for cotangent is cos/sin. So the asymptotes are located where sin (y) is equal to 0.Y is equal to 0 at 0 and pi, so this is the location of the asymptotes. We know that on the unit circle, Q1 begins at 0 and ends at pi/2.To find out how the graph for cot looks, we can use the trig ratio for cot: cos/sin. Both sin and cos are pos. in the first quadrant, so when you divide one by the other, you end up with a pos. value for cot. For Q2, from pi/2 to pi, we know that sin is pos. and cos is neg. This means cot is neg. in the second Quadrant because neg/pos=neg. Q3 is from pi to 3pi/2, sin and cos are both neg. A neg/neg gives us a pos. value for tan. In Q4, from 3pi/2 to pi, sin is pos and cos is neg. A neg./pos. gives us a neg. value for cotangent in the 4th quadrant. Thus, the graph for cotangent is pos, neg, pos, neg. When we graph it within its period of 0 and pi, the graph will begin close to the asymptote above the x-axis. When it enters Q2 it will be negative so it will start going downhill. This pattern will continue for every cotangent graph.

https://www.desmos.com/calculator/hjts26gwst |

Secant?

The trig ratio for secant is 1/cos. So the asymptotes are located where cos (x) is equal to 0. X is equal to 0 at pi/2 and 3pi/2, so this is the location of the asymptotes. We know that on the unit circle, Q1 begins at 0 and ends at pi/2.To find out how the graph for sec looks like, we can use our knowledge of trig ratio of sec: 1/cos. Because cos is pos. in the first quadrant, sec is also positive. For Q2, from pi/2 to pi, we know that cos is neg. making sec neg. Q3 is from pi to 3pi/2, in which cos is neg. 1/neg gives us a pos. value for sec. In Q4, from 3pi/2 to pi, cos is pos so sec is also pos. because a 1/pos. gives us a pos. value for secant in the 4th quadrant. Thus, the graph for sec t is pos, neg, neg, pos. Notice that this is the same pattern for cos! When we graph sec, we know that it takes 2pi to have an entire period. Our graph will look like an upside down parabola below the x-axis in Q2 and 3, with he vertex touching with the cos graph. However, in Q4 and 1, the parabola is above the x-axis and shares it vertex with the top of the cos graph.

https://www.desmos.com/calculator/hjts26gwst |

Cosecant?

The trig ratio for cosecant is 1/sin. So the asymptotes are located where sin (y) is equal to 0. Y is equal to 0 at 0 and pi, so this is the location of the asymptotes. We know that on the unit circle, Q1 begins at 0 and ends at pi/2.To find out how the graph for csc looks like, we can use our knowledge of trig ratio of csc: 1/sin. Because sin is pos. in the first quadrant, csc is also positive. For Q2, from pi/2 to pi, we know that sin is pos. making csc pos. Q3 is from pi to 3pi/2, because sin is neg. 1/neg gives us a pos. value for csc. In Q4, from 3pi/2 to pi, sin is neg so 1/neg. gives us a neg. value for csc in the 4th quadrant. Thus, the graph for csc is pos, pos, neg, neg. Notice that this is the same pattern for sin and csc! When we graph it withing Q1 and 2, the graph will look like a parabola above the x-axis in which it shares its vertex with the sin graph. When it enters Q3 and 4 it will look like an upside down parabola below the x-axis, sharing its vertex with sin as well.

https://www.desmos.com/calculator/hjts26gwst |

## No comments:

## Post a Comment