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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.
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