27)+Derive+trigonometric+identities+(Quotient+and+Pythagorean)

=Deriving trigonometric identities (Quotient and Pythagorean)=


 * Intro**: When we were doing this section I enjoyed when we derived identities because it was interesting to see how you could manipulate trig expressions in multiple ways so that they helped you answer various problems. I chose this topic because knowing how to derive these identities proved crucial when a test came around. Deriving these types of identities is simple, yet it is very important to know how to derive them in case you ever forget what the identities are.


 * Description**: Trigonometric identities are expressions that can be substituted in a variety of situations, such as while verifying and solving trig equations.

**Quotient Identities:**
To understand quotient identities, you first have to understand the six trigonometric functions on the unit circle. The six trig functions are tangent, sine, cosine, cotangent, cosecant, and secant; written as **tan, sin, cos, cot, csc, and sec**. Tan, sin, and cos are the "main" trig functions, while cot, csc, and sec are their inverses respectively. For example: tanx is the same as (1/cotx) If you recall SOHCAHTOA, this states that in a right triangle, sinx=opposite/hypotenuse, cosx=adjacent/hypotenuse, and tanx=opposite/adjacent. In a unit circle, right triangles can be formed in infinite positions, but for now, let's focus on a right triangle in the Quadrant I with a 45 degree angle. (pictured below) As you can see, the x and y values are labeled, and the hypotenuse is designated by the variable "r". Since sine is equal to the opposite side of the triangle divided by the hypotenuse, you can state that sinx=y/r, where x is the 45 degree angle at the origin. Similarly, you can state that cosx=x/r and tanx=y/x. Also, cscx=r/y, secx=r/x, and cotx=x/y. The quotient identities all come from these definitions of the trig functions.

The point of identities is to use them to substitute for something else to help you simplify problems. For the first quotient identity we are going to write tanx in a different way, by using two other trig functions to express it. tanx= ? Recall that tanx is the same as y/x. We need to find two trig functions that simplify to y/x when put together. Since sinx=y/r, and cosx=x/r, can you see how dividing them will cancel out the "r" variable? This leaves you with sinx / cosx = y/x Therefore, **tanx = sinx / cosx** This is a quotient identity, and either side of the relationship can be substituted for the other. Also from this, you can determine that **cotx = cosx / sinx** because cotangent is the inverse of tangent, and therefore, you can just flip the numerator and denominator from the previous identity.
 * Quotient Identity #1 and 2**
 * sinx / cosx = (y/r) / (x/r)
 * If you multiply by the reciprocal, this becomes sinx / cosx = (y/r) * (r/x)
 * Since one "r" is in the denominator, and the other is in the numerator, the two "r" variables disappear.

sinx= ? Using the same logic, sinx, which is equal to y/r, can be also written as (cosx/cotx) Therefore, **sinx = cosx / cotx** and **cscx = cotx / cosx**
 * Quotient Identity #3 and 4**
 * cosx=(x/r) and cotx=(x/y)
 * (x/r) / (x/y)
 * Multiply by reciprocal: (x/r) * (y/x)
 * "x" variables cancel and you are left with (y/r), which is the same as sinx.

cosx= ? Once again, using the same logic, cosx, which is equal to x/r, can be also written as (sinx/tanx) Therefore, **cosx = sinx / tanx** and **secx = tanx / sinx**
 * Quotient Identity #5 and 6**
 * sinx=(y/r) and tanx=(y/x)
 * (y/r) / (y/x)
 * Multiply by reciprocal: (y/r) * (x/y)
 * "y" variables cancel and you are left with (x/r), which is the same as cosx.

**Summary:**
 * The 6 Quotient Identities are:**
 * **tanx = sinx / cosx**
 * **cotx = cosx / sinx**
 * **sinx = cosx / cotx**
 * **cscx = cotx / cosx**
 * **cosx = sinx / tanx**
 * **secx = tanx / sinx**

**Pythagorean Identities:**
Pythagorean Identities also have the same use as quotient identities; to simplify a problem. As you might have guessed, these identities stem from the Pythagorean theorem, which states that in a right triangle, **a^2 + b^2 = c^2**. However, in this case, instead of using the variables "a", "b", and "c"... we use "x", "y", and "r", like we did during the quotient identities. Subsequently, we can state **x^2 + y^2 = r^2**. If you consider the unit circle, the radius is always going to be 1 unit. Therefore, the hypotenuse designated by the variable "r" can be replaced with a 1. So, the equation becomes x^2 + y^2 = 1^2, or more simply; ** x^2 + y^2 = 1. **

To find the first Pythagorean identity, we consider the expression ** sin ****^2****x + cos****^2****x**. As we discussed earlier, this is the same as **(y/r)^2 + (x/r)^2**. Squaring a fraction in parentheses is the same as squaring both the numerator and denominator, so you can write the above expression as: (y^2/r^2) + (x^2/r^2). As you can see, there are similar denominators, so you can combine the numerators like this: (x^2 + y^2) / r^2. Like we already stated, the variable "r" can be replaced by a 1. So, now our expression can be shown as (x^2 + y^2) / 1. The 1 in the denominator does not make a difference, so consequently, we are left with just x^2 + y^2. We already proved x^2 + y^2 = 1 by consulting the unit circle. Recall that we began with sin^2x and cos^2x, yet we simplified this down to x^2 + y^2, and found it equaled 1. Therefore, **sin****^2****x + cos****^2** ** x = 1 .** This is our first Pythagorean identity.
 * Pythagorean Identity #1**

To find the other two Pythagorean identities, we play around with the one we just found.

Start with the previous identity, **sin****^2****x + cos****^2****x = 1.** Divide each side by sin^2x. This becomes 1 + (cos^2x/sin^2x) = 1/sin^2x Recall the inverse of sinx, and also the quotient identities we just learned. Now the equation is 1 + (cot^2x) = csc^2x More commonly, the identity is shown equal to 1, so we subtract cot^2x from both sides. Our second identity is **csc^2x - cot^2x = 1**.
 * Pythagorean Identity #2**

Again, you begin with **sin****^2****x + cos****^2****x = 1**, but this time you divide both sides by cos^2x instead. Using inverses and the quotient identity sinx/cosx = tanx, the equation becomes tan^2x + 1 = sec^2x. Set equal to 1, the identity looks like **1 = sec^2x - tan^2x**.
 * Pythagorean Identity #3 **


 * Summary: **
 * The 3 Pythagorean Identities are: **
 * **sin****^2****x + cos****^2** ** x = 1 **
 * ** csc^2x - cot^2x = 1 **
 * ** sec^2x - tan^2x = 1 **

Practice Problems: Applying Quotient and Pythagorean Identities:
**1. Prove 2(sin^2x + cos^2x) = 2** Divide by 2 on both sides: sin^2x + cos^2x = 1 Substitute: **1=1**

Substitute sinx/cosx for tanx, and sinx/tanx for cosx. Multiply together: (sinx/cosx)(sinx/tanx) Simplify: (sin^2x/cosxtanx)
 * 2. Prove tanx * cosx = sin^2x/cosxtanx **
 * sin^2x/cosxtanx = sin^2x/cosxtanx**

**Class Problem:**
** Without checking the explanation, derive the three Pythagorean identities, beginning with the expression sin^2x + **** cos^2x ** **VIDEO:** []