IDNLearn.com offers a collaborative platform for sharing and gaining knowledge. Whether it's a simple query or a complex problem, our community has the answers you need.
Sagot :
Let's prove the given trigonometric identities in a step-by-step manner.
### Proving the First Identity
[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \cot \frac{B - C}{2} \][/tex]
To prove this, we use sum-to-product identities.
1. Sum-to-Product for Cosine:
[tex]\[ \cos B + \cos C = 2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) \][/tex]
2. Sum-to-Product for Sine:
[tex]\[ \sin B - \sin C = 2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right) \][/tex]
Now, substituting these identities into our original fraction:
[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \frac{2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)}{2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)} \][/tex]
3. Simplify the expression by canceling common factors:
[tex]\[ \frac{2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)}{2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)} = \frac{\cos \left( \frac{B - C}{2} \right)}{\sin \left( \frac{B - C}{2} \right)} \][/tex]
4. Recognize the quotient as cotangent:
[tex]\[ \frac{\cos \left( \frac{B - C}{2} \right)}{\sin \left( \frac{B - C}{2} \right)} = \cot \left( \frac{B - C}{2} \right) \][/tex]
Thus, we have proved the first identity:
[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \cot \frac{B - C}{2} \][/tex]
### Proving the Second Identity
[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = -\tan \frac{B - C}{2} \][/tex]
Similarly, we use sum-to-product identities for the second proof.
1. Sum-to-Product for Cosine:
[tex]\[ \cos B - \cos C = -2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right) \][/tex]
2. Sum-to-Product for Sine:
[tex]\[ \sin B + \sin C = 2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) \][/tex]
Now, substituting these identities into our original fraction:
[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = \frac{-2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)}{2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)} \][/tex]
3. Simplify the expression by canceling common factors:
[tex]\[ \frac{-2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)}{2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)} = \frac{- \sin \left( \frac{B - C}{2} \right)}{\cos \left( \frac{B - C}{2} \right)} \][/tex]
4. Recognize the quotient as tangent:
[tex]\[ \frac{- \sin \left( \frac{B - C}{2} \right)}{\cos \left( \frac{B - C}{2} \right)} = -\tan \left( \frac{B - C}{2} \right) \][/tex]
Thus, we have proved the second identity:
[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = -\tan \frac{B - C}{2} \][/tex]
Both identities are now proven.
### Proving the First Identity
[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \cot \frac{B - C}{2} \][/tex]
To prove this, we use sum-to-product identities.
1. Sum-to-Product for Cosine:
[tex]\[ \cos B + \cos C = 2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) \][/tex]
2. Sum-to-Product for Sine:
[tex]\[ \sin B - \sin C = 2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right) \][/tex]
Now, substituting these identities into our original fraction:
[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \frac{2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)}{2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)} \][/tex]
3. Simplify the expression by canceling common factors:
[tex]\[ \frac{2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)}{2 \cos \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)} = \frac{\cos \left( \frac{B - C}{2} \right)}{\sin \left( \frac{B - C}{2} \right)} \][/tex]
4. Recognize the quotient as cotangent:
[tex]\[ \frac{\cos \left( \frac{B - C}{2} \right)}{\sin \left( \frac{B - C}{2} \right)} = \cot \left( \frac{B - C}{2} \right) \][/tex]
Thus, we have proved the first identity:
[tex]\[ \frac{\cos B + \cos C}{\sin B - \sin C} = \cot \frac{B - C}{2} \][/tex]
### Proving the Second Identity
[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = -\tan \frac{B - C}{2} \][/tex]
Similarly, we use sum-to-product identities for the second proof.
1. Sum-to-Product for Cosine:
[tex]\[ \cos B - \cos C = -2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right) \][/tex]
2. Sum-to-Product for Sine:
[tex]\[ \sin B + \sin C = 2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) \][/tex]
Now, substituting these identities into our original fraction:
[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = \frac{-2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)}{2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)} \][/tex]
3. Simplify the expression by canceling common factors:
[tex]\[ \frac{-2 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right)}{2 \sin \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right)} = \frac{- \sin \left( \frac{B - C}{2} \right)}{\cos \left( \frac{B - C}{2} \right)} \][/tex]
4. Recognize the quotient as tangent:
[tex]\[ \frac{- \sin \left( \frac{B - C}{2} \right)}{\cos \left( \frac{B - C}{2} \right)} = -\tan \left( \frac{B - C}{2} \right) \][/tex]
Thus, we have proved the second identity:
[tex]\[ \frac{\cos B - \cos C}{\sin B + \sin C} = -\tan \frac{B - C}{2} \][/tex]
Both identities are now proven.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
