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NV
15 tháng 2 2019

Áp dụng công thức biến tích thành tổng:

\(cos\left(a+b\right).cos\left(a-b\right)=\dfrac{1}{2}\left(cos2a+cos2b\right)\)

\(=\dfrac{1}{2}\left(2cos^2a-1+1-2sin^2b\right)=\dfrac{1}{2}\left(2cos^2a-2sin^2b\right)\)

\(=cos^2a-sin^2b\)

\(cos\left(\dfrac{\pi}{4}+a\right).cos\left(\dfrac{\pi}{4}-a\right)+\dfrac{1}{2}sin^2a=\dfrac{1}{2}\left(cos\dfrac{\pi}{2}+cos2a\right)+\dfrac{1}{2}sin^2a\)

\(=\dfrac{1}{2}cos2a+\dfrac{1}{2}sin^2a=\dfrac{1}{2}\left(cos^2a-sin^2a\right)+\dfrac{1}{2}sin^2a\)

\(=\dfrac{1}{2}cos^2a\)

NV
13 tháng 4 2021

1.

\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)

\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)

\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)

\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)

\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)

\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)

\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)

\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)

NV
13 tháng 4 2021

2.

\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)

\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x\)

\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)

\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)

\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)

\(=tan^2a.cot^2b-2\)

25 tháng 4 2022

\(=\sin^2x+\dfrac{1}{2}\left[\cos\dfrac{2\Pi}{3}+\cos\left(-2x\right)\right]=\)

\(=\sin^2x+\dfrac{1}{2}\left(-\dfrac{1}{2}+\cos^2x-\sin^2x\right)=\)

\(=\sin^2x-\dfrac{1}{4}+\dfrac{1}{2}\cos^2x-\dfrac{1}{2}\sin^2x=\)

\(=\dfrac{1}{2}\left(\sin^2x+\cos^2x\right)-\dfrac{1}{4}=\dfrac{1}{4}\left(dpcm\right)\)

27 tháng 5 2021

A\(=\dfrac{cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}}{cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{7}.cos\dfrac{5\pi}{7}}\)

Đặt tử là Y; mẫu là U

Có \(Y=\)\(cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+\left(cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}\right)\)

\(=cos\left(\pi-\dfrac{2\pi}{7}\right).cos\left(\pi-\dfrac{4\pi}{7}\right)+cos\dfrac{\pi}{7}\left(cos\dfrac{5\pi}{7}+cos\dfrac{3\pi}{7}\right)\)

\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{\pi}{7}.2cos\dfrac{4\pi}{7}.cos\dfrac{\pi}{7}\)\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+2.cos^2\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}\)

\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+\left(cos\dfrac{2\pi}{7}+1\right).cos\dfrac{4\pi}{7}\)\(=2.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{4\pi}{7}\)

\(=cos\dfrac{6\pi}{7}+cos\dfrac{2\pi}{7}+cos\dfrac{4\pi}{7}\)

\(\Rightarrow sin\dfrac{\pi}{7}.Y=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}\left(-sin\dfrac{\pi}{7}+sin\dfrac{3\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{3\pi}{7}+sin\dfrac{5\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{5\pi}{7}+sin\pi\right)\)

\(=\dfrac{1}{2}\left(sin\pi-sin\dfrac{\pi}{7}\right)\)\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\)

\(\Rightarrow Y=-\dfrac{1}{2}\)

Có \(sin\dfrac{\pi}{7}.U=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{5}.cos\dfrac{5\pi}{7}\)

\(=\dfrac{1}{2}.sin\dfrac{2\pi}{7}.cos\left(\pi-\dfrac{2\pi}{7}\right).cos\dfrac{3\pi}{5}\)

\(=-\dfrac{1}{4}.sin\dfrac{4\pi}{7}.cos\left(\pi-\dfrac{4\pi}{5}\right)\)

\(=\dfrac{1}{8}.sin\dfrac{8\pi}{7}\)\(=\dfrac{1}{8}.sin\left(\pi+\dfrac{\pi}{7}\right)=-\dfrac{1}{8}.sin\dfrac{\pi}{7}\)

\(\Rightarrow U=-\dfrac{1}{8}\) 

Vậy \(A=\dfrac{Y}{U}=4\)

27 tháng 5 2021

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CHÚC BẠN HỌC TỐT NHÉok

AH
Akai Haruma
Giáo viên
28 tháng 5 2018

Lời giải:

Do \(0< a< \frac{\pi}{2}\Rightarrow \sin a>0\)

Ta có:

\(\sqrt{\frac{1+\cos a}{1-\cos a}}-\sqrt{\frac{1-\cos a}{1+\cos a}}=\frac{(1+\cos a)-(1-\cos a)}{\sqrt{(1-\cos a)(1+\cos a)}}\)

\(=\frac{2\cos a}{\sqrt{1-\cos ^2a}}=\frac{2\cos a}{\sqrt{\sin ^2a}}=\frac{2\cos a}{\sin a}\)

\(=2.\frac{\cos a}{\sin a}=2\cot a\)

Ta có đpcm.

3 tháng 6 2021

\(cos\dfrac{\pi}{5}-cos\dfrac{2\pi}{5}\)

\(=-2.sin\dfrac{3\pi}{10}.sin\left(-\dfrac{\pi}{10}\right)\)

\(=2.sin\left(\dfrac{1}{2}-\dfrac{\pi}{5}\right).sin\dfrac{\pi}{10}\)

\(=2.sin\dfrac{\pi}{10}.cos\dfrac{\pi}{5}=\dfrac{sin\dfrac{\pi}{5}.cos\dfrac{\pi}{5}}{cos\dfrac{\pi}{10}}\)

\(=\dfrac{\dfrac{1}{2}sin\dfrac{2\pi}{5}}{cos\left(\dfrac{\pi}{2}-\dfrac{2\pi}{5}\right)}=\dfrac{\dfrac{1}{2}.sin\dfrac{2\pi}{5}}{sin\dfrac{2\pi}{5}}\)\(=\dfrac{1}{2}\)

cosπ5−cos2π5cosπ5−cos2π5

=−2.sin3π10.sin(−π10)=−2.sin3π10.sin(−π10)

=2.sin(12−π5).sinπ10=2.sin(12−π5).sinπ10

=2.sinπ10.cosπ5=sinπ5.cosπ5cosπ10=2.sinπ10.cosπ5=sinπ5.cosπ5cosπ10

=12sin2π5cos(π2−2π5)=12.sin2π5sin2π5=12sin2π5cos(π2−2π5)=12.sin2π5sin2π5=12

18 tháng 4 2017

\(\cos\dfrac{\pi}{15}.\cos\dfrac{2\pi}{15}...\cos\dfrac{7\pi}{15}=-\dfrac{1}{2}.\left(\cos\dfrac{\pi}{15}.\cos\dfrac{2\pi}{15}.\cos\dfrac{4\pi}{15}.\cos\dfrac{8\pi}{15}\right).\left(\cos\dfrac{3\pi}{15}.\cos\dfrac{6\pi}{15}\right)\)

\(=-\dfrac{1}{2}.\left(\cos\dfrac{\pi}{15}.\cos\left(2.\dfrac{\pi}{15}\right).\cos\left(2^2.\dfrac{\pi}{15}\right).\cos\left(2^3\dfrac{\pi}{15}\right)\right).\left(\cos\dfrac{3\pi}{15}.\cos\left(2.\dfrac{3\pi}{15}\right)\right)\)

\(=-\dfrac{1}{2}.\left(\dfrac{\sin\left(2^4.\dfrac{\pi}{15}\right)}{16.\sin\left(\dfrac{\pi}{15}\right)}\right).\left(\dfrac{\sin\left(2^2\dfrac{3\pi}{15}\right)}{4.\sin\left(\dfrac{3\pi}{15}\right)}\right)\)

\(=-\dfrac{1}{2}.\left(\dfrac{\sin\left(\dfrac{16\pi}{15}\right)}{16.\sin\left(\dfrac{\pi}{15}\right)}\right).\left(\dfrac{\sin\left(\dfrac{12\pi}{15}\right)}{4.\sin\left(\dfrac{3\pi}{15}\right)}\right)\)

\(=-\dfrac{1}{2}.\left(\dfrac{-\sin\left(\dfrac{\pi}{15}\right)}{16.\sin\left(\dfrac{\pi}{15}\right)}\right).\left(\dfrac{\sin\left(\dfrac{3\pi}{15}\right)}{4.\sin\left(\dfrac{3\pi}{15}\right)}\right)=\dfrac{1}{128}\)