Establish each identity. (1+sinθ+cosθ)/(1+sinθ-cosθ)=(1+cosθ)/(sinθ)

Name: Problem 87, Chapter 7: Precalculus: pearson new international edition
Uploaded: 2020-06-20T01:23:10-08:00
Duration: 12 min 25 s
Channel: Peter Winans
Description: Establish each identity. (1+sinθ+cosθ)/(1+sinθ-cosθ)=(1+cosθ)/(sinθ)

step 1 All right. We need to prove that this identity is true. In other words, we need to show that the

Establish each identity....

Key Concepts

Simplification of Complex Fractions

Simplifying complex fractions involves rewriting the expression to cancel common factors and combine terms effectively. This often requires careful rearrangement of numerator and denominator expressions while ensuring that the operations conform to the rules of algebra. Such simplification is a key step in verifying that both sides of a trigonometric identity are indeed equal.

Algebraic Manipulation

Algebraic manipulation in trigonometry includes techniques such as factoring, expanding expressions, combining like terms, and using common denominators. It is crucial for transforming complex expressions into a more workable form so that the equivalence between the sides of a trigonometric identity can be clearly seen.

Verification of Trigonometric Identities

This concept involves demonstrating that two seemingly different trigonometric expressions are equivalent. It requires a systematic approach of manipulating one or both sides of an equation using known trigonometric identities, algebraic techniques, and logical reasoning. This process is fundamental in trigonometry because it reinforces the inherent relationships between trigonometric functions.

Transcript

00:03 All right.

00:05 We need to prove that this identity is true.

00:07 In other words, we need to show that the left side and the right side are the same.

00:11 We're going to start by working on the left side.

00:15 That seems to be the more complicated side.

00:17 And i need to get this so that it's equal to the sine of theta plus one over the cosine of theta.

00:24 So i'm going to start by doing something a little bit unusual.

00:27 It's not like a lot of the other proofs that we've seen so far in this section.

00:34 I'm going to have the sine of theta.

00:37 Now i'm going to group the second two terms together in my numerator.

00:40 I'm also going to factor out that negative sign right there.

00:44 So i'm going to have minus the quantity, cosine of theta minus 1.

00:51 If i factor out a negative 1 from negative cosine, i get cosine, negative 1 from 1.

00:55 I get negative 1.

00:57 And in my denominator, i will have the sine of theta.

01:02 And if i group those second two terms together, i'm simply going to get cosine of theta minus 1.

01:09 Now, in that term right there, i have sine of theta minus cosine of theta minus 1 over sine theta plus cosine theta minus 1.

01:24 We want to think about this denominator is the difference of two terms, right? the sine of theta is one term, and the cosine of theta minus 1 is the other term.

01:34 What we're going to do next is multiply this thing, this fraction, by the conjugate of the denominator.

01:46 In other words, i'm going to multiply by sine of theta minus cosine of theta minus 1 over sine of theta minus 1.

02:10 And the reason i'm able to do that is because i'm multiplied by something over itself, it's got to be equal to 1.

02:17 So i'm technically only multiplying by 1 here, which is the multiplicative identity, which means we're not changing the value of this.

02:23 When i multiply my two numerators, i have sine of theta minus, cosine of theta minus 1, times sine of theta minus the cosine of theta minus 1.

02:37 I'm multiplying something by itself, in other words, so i'm squaring it.

02:42 So when you square a binomial, you start by squaring the first term.

02:51 So that's going to give a sine squared of theta minus, then you multiply the two terms by each other and double it.

02:57 So it's going to be minus 2, sine theta, times cosine of theta minus 1.

03:06 And then you squared the second term, so this is going to be plus cosine of theta minus 1 squared.

03:17 And this is all going to be over.

03:21 When you multiply conjugate pairs, the middle terms always add to zero.

03:26 So we get sine squared of theta minus cosine, cosine, of theta minus 1 squared.

03:40 So the next thing we're going to need to do then is square those two binomials.

03:47 And we are going to get, when we do so, sine squared of theta minus two sine of theta times cosine of theta minus one, plus, now when i square cosine theta minus one here, i'm going to get cosine squared theta minus, 2, cosine of theta plus 1.

04:26 And that's all going to be over the sine squared of theta minus.

04:36 Now, i'm squaring this binomial, but i'm also subtracting it, so i'm going to need to keep it in parentheses here, so i remember to distribute the negative sign.

04:44 We are going to get the cosine squared of theta minus 2 cosine of theta plus 1...

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