Chapter+9

=
Chapter 9: // Analytic Trigonometry // ===== = =

===This chapter highlights the uses of trigonometric identities in determining the values of expressions involving such identities as well as applying them to solving equations. flat ===

= Chapter 9 Section 4 =

// __Sum and Difference Formulas__ //  If you are able to find the sum or difference of two trigonometric functions, you can determine the six functions of almost any angle, utilizing previous knowledge of angles such as 30 °, 45 ° , 60 ° , and quadrantal angles. For example, although you many not readily know the sine of 75 °, you can determine the value by adding the sine values of 30 ° and 45 °. The same holds true for cosine and tangent. However, there are different formulas for each function. The sums and differences of reciprocal functions; cosecant, secant, and cotangent, can be found by evaluating the expression in terms of sine, cosine, or tangent, and simply taking the reciprocal of the result.   ** Finding the sum of tangents: ** ** Finding the difference of tangents: ***Helpful video for sum and difference of sine, cosine, and tangent media type="youtube" key="MhdriAGLfa0" height="385" width="480" = =

Examples
= =

"Working Backwards"
If you are given an expression such as, you can find the value by determining what sum or difference identity is being utilized. In this case, the difference of two sine functions is being shown, the result being 60 degrees. You can find the sine of 60 degrees,, with the 30, 60, 90 triangle dimensions. = =

Establishing Trigonometric Identities
Since sine and cosecant are reciprocal functions, we can establish that the cosecant squared of angle y is equal to: When multiplied by sine of angle y 1 sine of angle y cancels and the result is: Which, because of its reciprocality,can also be expressed as the right side of the original expression,



//**A helpful resource for practice problems is**// //**Interactive Mathematics**// //**. This website provides great examples with detailed solutions upon request.**//

Another Example
** Prove that ** · ** Use the difference formula of cosine ** cosαosβ+sinαsinβ/sinαsinβ=cotαotβ+1 · ** Separate the left side at the addition sign ** cosαcosβ/sinαsinβ+ sinαsinβ/sinαsinβ =cotαotβ+1 · ** Change sinαsinβ/sinαsinβ into 1, because something over itself is 1 ** cosαcosβ/sinαsinβ +1= cotαotβ+1 · ** Separate the fraction ** cosα/sinα· cosβ/sinβ+1= cotαotβ+1 · ** Use reciprocal functions to change the fractions
 * // cos(α-β)/sinαsinβ=cotαcotβ+1 //
 * cotαcotβ+1= cotαotβ+1 **

> = Chapter 9 Section 5 =

** __//Double-Angle and Half-Angle Formulas//__ In this section, we examine the ways that you can determine the value of a trigonometric function for an angle that has been doubled, or halved. In this way, you can more easily determine the values of expressions such as the cosine of 15 degrees, or the tangent of 120 degrees. It requires the memorization of several formulas, however, once you learn them, analyzing any trigonometric expression will become much easier.  __Double-Angle Formulas__ ** __** Helpful video for Double-Angle Formulas **__

**  media type="youtube" key="mP6wujhQ8js" height="385" width="480"

__Variations of Double-Angle Formulas__ __Half-Angle Formulas__ If we let theta equal alpha over two as shown above, these can also be written as:

Another half-angle formula for tangent can be found by taking  and rewriting it as __Helpful video for Half-Angle Formulas__ media type="youtube" key="__7uCEE5iV8" height="385" width="480"

** ** Examples **

<span style="background-color: #d02f2f; display: block; font-size: 24px; line-height: 35px; text-align: center;">**Establishing Trigonometric Identities**
 * // Example //**

· ** Prove that ** cot(2θ) = (cot2 θ − 1) ⁄ (2cotθ) · ** Use reciprocal to change cot(2 **** θ **** ) ** cot (2θ) = 1 ⁄ tan(2θ) · ** Use Double-Angle Formulas for tan(2 **** θ **** ) ** cot(2θ) = 1⁄ (2tanθ ⁄ 1 − tan2 θ) · ** Solve 1 **** ⁄ **** (2tan **** θ ⁄ 1 − tan **** 2 **** θ **** ) ** cot(2θ) = (1− tan2 θ) ⁄ (2tanθ) · ** Use reciprocal for tan **** 2 ** ** θ and tanθ ** cot(2θ) = (1 − 1 ⁄ cot2 θ) ⁄ (2 ⁄ cotθ) · ** Solve for (1 − 1 ⁄ cot **** 2 **** θ) ** cot(2θ) = (cot2 θ −1 ⁄ cot2 θ) ⁄ (2 ⁄ cotθ) · ** Solve for **** (cot **** 2 **** θ ** ** − **** 1 **** ⁄ **** cot2 ** ** θ **** ) **** ⁄ **** (2 **** ⁄ **** cot **** θ **** ) ** cot(2θ) = (cot2 θ − 1) ⁄ (2cotθ)