Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Alright, let's break down the given equation step-by-step to prove that the left-hand side is equal to the right-hand side:
[tex]\[ \frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \tan (A+B) \][/tex]
### Step 1: Use the Difference of Squares Identity
We start by simplifying the numerator using the difference of squares identity:
[tex]\[ \sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B) \][/tex]
### Step 2: Simplify the Denominator Using Angle Addition and Subtraction Formulas
To simplify the term [tex]\(\sin A \cdot \cos A - \sin B \cdot \cos B\)[/tex], we can use the angle addition and subtraction formulas:
[tex]\[ \sin A \cos A = \frac{1}{2} \sin(2A) \quad \text{and} \quad \sin B \cos B = \frac{1}{2} \sin(2B) \][/tex]
Thus, the denominator becomes:
[tex]\[ \sin A \cdot \cos A - \sin B \cdot \cos B = \frac{1}{2} \sin(2A) - \frac{1}{2} \sin(2B) = \frac{1}{2} (\sin(2A) - \sin(2B)) \][/tex]
### Step 3: Use Sum-to-Product Identities
Next, we use the sum-to-product identities:
[tex]\[ \sin x - \sin y = 2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right) \][/tex]
Applying this to our expression:
[tex]\[ \sin(2A) - \sin(2B) = 2 \cos\left(A + B\right) \sin\left(A - B\right) \][/tex]
Thus:
[tex]\[ \frac{1}{2} (\sin(2A) - \sin(2B)) = \frac{1}{2} \cdot 2 \cos(A + B) \sin(A - B) = \cos(A + B) \sin(A - B) \][/tex]
### Step 4: Substitute These into the Original Fraction
Now we substitute the simplified numerator and denominator back into the original fraction:
[tex]\[ \frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A + B) \sin(A - B)} \][/tex]
### Step 5: Cancel the Common Factor
Notice that [tex]\(\sin(A - B)\)[/tex] is a common factor in the numerator and denominator:
[tex]\[ \frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A + B) \sin(A - B)} = \frac{\sin A + \sin B}{\cos(A + B)} \][/tex]
### Step 6: Recognize the Tangent Function
Recall the definition of the tangent function:
[tex]\[ \tan (A+B) = \frac{\sin (A+B)}{\cos (A+B)} \][/tex]
We can recognize that:
[tex]\[ \frac{\sin(\alpha) + \sin(\beta)}{\cos (\alpha + \beta)} \Rightarrow \sin\alpha = \sin A \quad \text{and} \quad \sin\beta = \sin B, i.e., \alpha = A \quad \text{and} \quad \beta = B \][/tex]
Thus:
[tex]\[ \frac{\sin(A + B)}{\cos(A + B)} = \tan(A + B) \][/tex]
Therefore, the given equation simplifies to:
[tex]\[ \frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \tan(A+B) \][/tex]
So, the equation is an identity and holds true.
[tex]\[ \frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \tan (A+B) \][/tex]
### Step 1: Use the Difference of Squares Identity
We start by simplifying the numerator using the difference of squares identity:
[tex]\[ \sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B) \][/tex]
### Step 2: Simplify the Denominator Using Angle Addition and Subtraction Formulas
To simplify the term [tex]\(\sin A \cdot \cos A - \sin B \cdot \cos B\)[/tex], we can use the angle addition and subtraction formulas:
[tex]\[ \sin A \cos A = \frac{1}{2} \sin(2A) \quad \text{and} \quad \sin B \cos B = \frac{1}{2} \sin(2B) \][/tex]
Thus, the denominator becomes:
[tex]\[ \sin A \cdot \cos A - \sin B \cdot \cos B = \frac{1}{2} \sin(2A) - \frac{1}{2} \sin(2B) = \frac{1}{2} (\sin(2A) - \sin(2B)) \][/tex]
### Step 3: Use Sum-to-Product Identities
Next, we use the sum-to-product identities:
[tex]\[ \sin x - \sin y = 2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right) \][/tex]
Applying this to our expression:
[tex]\[ \sin(2A) - \sin(2B) = 2 \cos\left(A + B\right) \sin\left(A - B\right) \][/tex]
Thus:
[tex]\[ \frac{1}{2} (\sin(2A) - \sin(2B)) = \frac{1}{2} \cdot 2 \cos(A + B) \sin(A - B) = \cos(A + B) \sin(A - B) \][/tex]
### Step 4: Substitute These into the Original Fraction
Now we substitute the simplified numerator and denominator back into the original fraction:
[tex]\[ \frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A + B) \sin(A - B)} \][/tex]
### Step 5: Cancel the Common Factor
Notice that [tex]\(\sin(A - B)\)[/tex] is a common factor in the numerator and denominator:
[tex]\[ \frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A + B) \sin(A - B)} = \frac{\sin A + \sin B}{\cos(A + B)} \][/tex]
### Step 6: Recognize the Tangent Function
Recall the definition of the tangent function:
[tex]\[ \tan (A+B) = \frac{\sin (A+B)}{\cos (A+B)} \][/tex]
We can recognize that:
[tex]\[ \frac{\sin(\alpha) + \sin(\beta)}{\cos (\alpha + \beta)} \Rightarrow \sin\alpha = \sin A \quad \text{and} \quad \sin\beta = \sin B, i.e., \alpha = A \quad \text{and} \quad \beta = B \][/tex]
Thus:
[tex]\[ \frac{\sin(A + B)}{\cos(A + B)} = \tan(A + B) \][/tex]
Therefore, the given equation simplifies to:
[tex]\[ \frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \tan(A+B) \][/tex]
So, the equation is an identity and holds true.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.