Question

ゆめら

（I）の問題でどうしてx＝7／4πのときに最大値√2をとるのかわかりません。

308 次の関数の最大値、最小値を求めよ。 (1), (2)について
値も求めよ。
(1) y=-sinx+cosx (0<x<27)
*(2) y=sinx+√3c
cosx (0≤x≤t)
(3) y=√7 sinx-3cosx
*(4) y=2sinx+cosx (0≤x≤t)
浸優先

(1) −sinx+cosx=V2 sin|x+
から
よって
また
3
y=√2 sin(x+³/7)
¹ ( x + 3/1/²
4
x)
よって
3
3
0≦x<2のとき,0242x+2/ <1/12 である
π
↑
4
3
-1sin(x+2/27) 201
≤1
-√2≦x≦√2
-√√2 ≤y≤√²6 Y
3
sin x+ ・π
sin(x+2)=1のとき 子
x=
π
1 + b =
t=-1のとき
Kin
4
したがって,この関数は
a
7
x=²で最大値√2 をとり,
4
3
x= -で最小値-√2 をとる。
4
bad
L
2
TC
IL
2
x=7708
2
4

Answer 1

✨ ベストアンサー ✨

Dw

Use Angle sum and difference identities.
y=√2sin(x+3/4π)
Under normal circumstances,
-1≤sin(any)≤1
-1≤sin(x+3/4π)≤1
-√2≤√2sin(x+3/4π)≤√2
To achieve the maximum value, the sine function naturally needs to be at its maximum, which is equal to 1. In order for the sine value to be equal to 1, theta must be π/2 + 2nπ.
In this case, 3/4π+7/4π=10/4π=π/2+2π→sin(x+3/4π)=1
√2×1=√2,x=7/4π

ゆめら

Why does θ need to be π/2＋2nπ for the sine value to be equal to 1？
Is 2nπ necessary ？

Sorry if I said something off topic

Dw

No need to apologize. Don't worry.👋
The expression 2nπ indicates an additional rotation of n circles.
When n=1, it signifies an extra rotation of one circle.
When n=2, it denotes an extra rotation of two circles, and so on. However, in this question, it might not be crucial. I just want to express the sine values at π/2 for each.

ゆめら

Thank you for your detailed explanation. I understand! thank you very much