Suppose that tanθ=p / q, where p and q are positive and 0^∘<θ<90^∘ . Show that (p sinθ-q cosθ)/(

Suppose that tanθ=p / q, where p and q are positive and...

Suppose that $\tan \theta=p / q,$ where $p$ and $q$ are positive and $0^{\circ}<\theta<90^{\circ} .$ Show that
$$
\frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}=\frac{p^{2}-q^{2}}{p^{2}+q^{2}}
$$

Key Concepts

Pythagorean Identity

The Pythagorean identity, sin²? + cos²? = 1, is essential in many trigonometric proofs and simplifications. In problems where the tangent function is given as a ratio of sides, this identity enables the normalization of sine and cosine so that their expressions are consistent with the identity, thus allowing the derivation of further relationships between the functions.

Algebraic Simplification

Algebraic manipulation plays a crucial role in trigonometry. It involves operations such as factoring, expanding, and simplifying complex expressions. In the context of these kinds of trigonometric proofs, algebraic simplification allows one to combine and reduce expressions, like ratios involving sine and cosine, to reveal underlying identities or equalities such as the given result.

Trigonometric Ratios

This concept pertains to understanding the functions sine, cosine, and tangent in the context of a right triangle. When tan ? is given as a ratio of two sides, it allows one to express the sine and cosine functions in terms of those sides, thereby linking the different trigonometric functions together. This is a fundamental tool in converting problems stated in one trigonometric function into forms that involve others.

Transcript

00:04 All right, we want to show in this problem that p times the sine of theta minus q times the cosine of theta divided by p times the sine of theta plus q times the cosine of theta is equal to p squared minus q squared over p squared plus q squared.

00:43 And the only real piece of information that we were given to work with is the fact that the tangent of theta is equal to p over q.

00:57 So we're going to have to start with that.

01:00 We'll come back to this equation in a minute, but we're going to have to start with this tangent of theta right here.

01:06 Because if i know tangent of theta is equal to p over q, i can draw some generic triangle here.

01:16 Remember tangent is opposite over adjacent.

01:22 So if this is angle theta, then the opposite side is p, the adjacent side is q.

01:29 In the equation that we have to play with, where you have to talk about the sign and the cosine of theta.

01:37 Well, to find both of those, i'm going to need the length of the hypotenuse.

01:42 So the first thing i really have to do here is use the pythagorean theorem to find, find the length of the hypotenuse.

01:51 And that's going to be h squared.

01:55 The hypotenuse squared is going to equal the sum of the square of the legs.

01:59 That's just going to be p squared plus q squared.

02:04 And now if i take the square root of both sides of this thing, h is going to be the square root of p squared plus q squared, which allows me to then talk about the sine of theta, which is the opposite its side over the hypotenuse...

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