Question

Kết quả nào sau đây sai

A. \(sinx+cosx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)

B. \(sinx-cosx=\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\)

C. \(cosx+sinx=\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)\)

D. \(cosx+sinx=\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)\)

Giải thích luôn ạ

Answer 1

a) \(A=sinx+cosx\)

\(\Leftrightarrow A\dfrac{\sqrt{2}}{2}=sinx.\dfrac{\sqrt{2}}{2}+cosx.\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow A\dfrac{\sqrt{2}}{2}=sinx.cos\dfrac{\pi}{4}+cosx.sincos\dfrac{\pi}{4}\)

\(\Leftrightarrow A\dfrac{1}{\sqrt{2}}=sin\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow A=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(đpcm\right)\)

b) \(A=sinx-cosx\)

\(\Leftrightarrow A\dfrac{\sqrt{2}}{2}=sinx.\dfrac{\sqrt{2}}{2}-cosx.\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow A\dfrac{\sqrt{2}}{2}=sinx.cos\dfrac{\pi}{4}-cosx.sincos\dfrac{\pi}{4}\)

\(\Leftrightarrow A\dfrac{1}{\sqrt{2}}=sin\left(x-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow A=\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\left(đpcm\right)\)

c) Theo câu a ta được :  \(cosx+sinx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)

d)  \(cosx+sinx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)

\(=\sqrt{2}sin\left(\dfrac{\pi}{2}+x-\dfrac{\pi}{4}\right)=\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)\left(đpcm\right)\)

 

 

 

 

Answer 2

Câu C 

Theo câu A

\(cosx+sinx=\sqrt[]{2}sin\left(x+\dfrac{x}{4}\right)=\sqrt[]{2}sin\left(\dfrac{x}{2}+x-\dfrac{x}{4}\right)\)

\(=\sqrt[]{2}cos\left(x-\dfrac{x}{4}\right)\)

Bạn xem lại đề câu C

Answer 3

Vậy chọn C là đáp án sai