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Sagot :
To express the product [tex]\(\sin(11x) \cos(x)\)[/tex] as a sum or difference of sines and/or cosines, we can make use of a specific trigonometric identity. The relevant identity is:
[tex]\[ \sin(A) \cos(B) = \frac{1}{2} \left[ \sin(A + B) + \sin(A - B) \right] \][/tex]
Here, [tex]\(A = 11x\)[/tex] and [tex]\(B = x\)[/tex].
Using the identity mentioned, we proceed as follows:
1. Identify [tex]\(A + B\)[/tex]:
[tex]\[ A + B = 11x + x = 12x \][/tex]
2. Identify [tex]\(A - B\)[/tex]:
[tex]\[ A - B = 11x - x = 10x \][/tex]
3. Substitute [tex]\(A + B\)[/tex] and [tex]\(A - B\)[/tex] back into the identity:
[tex]\[ \sin(11x) \cos(x) = \frac{1}{2} \left[ \sin(12x) + \sin(10x) \right] \][/tex]
Thus, the product [tex]\(\sin(11x) \cos(x)\)[/tex] can be written as:
[tex]\[ \sin(11x) \cos(x) = \frac{\sin(12x)}{2} + \frac{\sin(10x)}{2} \][/tex]
So, the final simplified answer is:
[tex]\[ \sin(11x) \cos(x) = \frac{1}{2} \sin(12x) + \frac{1}{2} \sin(10x) \][/tex]
[tex]\[ \sin(A) \cos(B) = \frac{1}{2} \left[ \sin(A + B) + \sin(A - B) \right] \][/tex]
Here, [tex]\(A = 11x\)[/tex] and [tex]\(B = x\)[/tex].
Using the identity mentioned, we proceed as follows:
1. Identify [tex]\(A + B\)[/tex]:
[tex]\[ A + B = 11x + x = 12x \][/tex]
2. Identify [tex]\(A - B\)[/tex]:
[tex]\[ A - B = 11x - x = 10x \][/tex]
3. Substitute [tex]\(A + B\)[/tex] and [tex]\(A - B\)[/tex] back into the identity:
[tex]\[ \sin(11x) \cos(x) = \frac{1}{2} \left[ \sin(12x) + \sin(10x) \right] \][/tex]
Thus, the product [tex]\(\sin(11x) \cos(x)\)[/tex] can be written as:
[tex]\[ \sin(11x) \cos(x) = \frac{\sin(12x)}{2} + \frac{\sin(10x)}{2} \][/tex]
So, the final simplified answer is:
[tex]\[ \sin(11x) \cos(x) = \frac{1}{2} \sin(12x) + \frac{1}{2} \sin(10x) \][/tex]
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