Chapter 11: Problem 270
$$ \text { If } \cos (A+B) \sin (C+D)=\cos (A-B) \sin (C-D), \text { prove that } \cot A \cot B \cot C=\cot D \text { . } $$
Short Answer
Expert verified
The proof is complete, the given equation \( \cos (A+B) \sin (C+D)=\cos (A-B) \sin (C-D) \) does imply that \( \cot A \cot B \cot C = \cot D \).
Step by step solution
01
Use the Co-Function Identities
Knowing that \( \sin (90° - x) = \cos x \) and \( \cos (90° - x) = \sin x \), replace \( \cos (A±B) \) by \( \sin (90° - (A±B)) \) and \( \sin (C±D) \) by \( \cos (90° - (C±D)) \) in the given equation.
02
Utilize the Complementary Angle Identity
Apply the Complementary Angle Identity \( \cos (90° - x) = \sin x \) and \( \sin (90° - x) = \cos x \). Now the equation reads \( \sin (90° - (A+B)) \cos (90° - (C+D)) = \sin (90° - (A-B)) \cos (90° - (C-D)) \) which can be simplified to \( \cos (A+B) \sin (C+D) = \cos (A-B) \sin (C-D) \).
03
Utilize the Reciprocal/TRIG Identities
Knowing that \( \cot x = \frac{1}{\tan x} \) and \( \tan x = \frac{\sin x}{\cos x} \), we can express each of \( \cot A \), \( \cot B \), \( \cot C \), and \( \cot D \) in terms of Sine and Cosine functions. This leads to \( \frac{\cos A}{\sin A} \frac{\cos B}{\sin B} \frac{\cos C}{\sin C}=\frac{\cos D}{\sin D} \). Rewriting the equation, we get: \( \cos A \cos B \cos C = \sin A \sin B \sin C \cdot \cot D \). Given that \( \cos A \cos B \cos C = \sin A \sin B \sin C \) from the initial expression derived in Step 2, we can substitute into the above equation, yielding \( \cot D = \cot A \cot B \cot C \). This is the required proof.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cotangent
Cotangent, often abbreviated as "cot," is a fundamental trigonometric function. It is the reciprocal of the tangent function. Mathematically, this is expressed as \( \cot x = \frac{1}{\tan x} \). To understand it better, we break it down to its sine and cosine components:
- The tangent function is defined as the ratio of sine to cosine: \( \tan x = \frac{\sin x}{\cos x} \)
- Thus, cotangent, being the reciprocal, becomes \( \cot x = \frac{\cos x}{\sin x} \)
Co-Function Identities
Co-function identities are a crucial part of trigonometry, helping us understand the relationships between different trigonometric functions. They are based on complementary angles, which add up to \( 90^\circ \) or \( \frac{\pi}{2} \) radians. For instance:
- The co-function identity for sine and cosine is \( \sin(90^\circ - x) = \cos x \) and \( \cos(90^\circ - x) = \sin x \)
- Similarly, for tangent and cotangent: \( \tan(90^\circ - x) = \cot x \) and \( \cot(90^\circ - x) = \tan x \)
Reciprocal Identities
Reciprocal identities are another set of fundamental relationships in trigonometry. They express trigonometric functions as reciprocals of one another. In simple terms:
- Cosecant is the reciprocal of sine: \( \csc x = \frac{1}{\sin x} \)
- Secant is the reciprocal of cosine: \( \sec x = \frac{1}{\cos x} \)
- Cotangent, as previously mentioned, is the reciprocal of tangent: \( \cot x = \frac{1}{\tan x} \)
Complementary Angle Identity
The complementary angle identity is an extension of the co-function identities and revolves around angles that complement each other—meaning their sum is \( 90^\circ \). This concept simplifies expressions by equating certain trigonometric functions directly:
- Given a set of angles \( A \) and \( B \), if \( A + B = 90^\circ \), then \( \sin A = \cos B \) and \( \cos A = \sin B \)