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1 個解答
- 匿名3 月前
Question：
Prove [sin x + sin(x + y) + sin(x + 2y)]/[cos x + cos(x + y) + cos(x + 2y)] = tan(x + y).
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Solution：
Suppose cos x + cos(x + y) + cos(x + 2y) ≠ 0.
[sin x + sin(x + y) + sin(x + 2y)]/[cos x + cos(x + y) + cos(x + 2y)] = tan(x + y)
sin x + sin(x + y) + sin(x + 2y) = tan(x + y) [cos x + cos(x + y) + cos(x + 2y)]
LHS
= [sin x + sin(x + 2y)] + sin(x + y)
= 2 sin[(x + x + 2y)/2] cos(2y/2) + sin(x + y)
= sin(x + y) (2 cos y + 1)
RHS
= tan(x + y) {[cos x + cos(x + 2y)] + cos(x + y)}
= tan(x + y) {2 cos[(x + x + 2y)/2] cos(2y/2) + cos(x + y)}
= tan(x + y) [2 cos(x + y) cos y + 1]
= sin(x + y) [cos(x + y) / cos(x + y)] (2 cos y + 1) ...... Assume cos(x + y) ≠ 0
= sin(x + y) (2 cos y + 1)
If cos(x + y) = 0, tan(x + y) is undefined.
∴ cos(x + y) ≠ 0.
∴ [sin x + sin(x + y) + sin(x + 2y)]/[cos x + cos(x + y) + cos(x + 2y)] = tan(x + y)
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Remarks：
1.
Sum to product formulas
sinA + sinB = 2 sin[(A + B)/2] cos[(A - B)/2]
cosA + cosB = 2 cos[(A + B)/2] cos[(A - B)/2]
2.
Here is the reason why I put "cos x + cos(x + y) + cos(x + 2y)" to the right hand side.
[sin x + sin(x + y) + sin(x + 2y)]/[cos x + cos(x + y) + cos(x + 2y)]
= [sin(x + y) (2 cos y + 1)]/[cos(x + y) (2 cos y + 1)]
= tan(x + y) [(2 cos y + 1)/(2 cos y + 1)] ...... Assume cos y ≠ -1/2
= tan(x + y)
When cos y = -1/2,
LHS
= ...
= RHS