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NV
29 tháng 5 2020

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

\(=2+2\left(cosa.cosb+sina.sinb\right)\)

\(=2+2.cos\left(a-b\right)=2+2.cos\frac{\pi}{3}=3\)

\(B=cos^2a+sin^2b+2cosa.sinb+cos^2b+sin^2a-2sina.cosb\)

\(=2-2\left(sina.cosb-cosa.sinb\right)\)

\(=2-2sin\left(a-b\right)=2-2sin\frac{\pi}{3}=2-\sqrt{3}\)

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.

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

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

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

\(=cos3x\)

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

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

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

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

19 tháng 8 2021

a, \(\dfrac{1-sin2a}{1+sin2a}\)

\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)

\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)

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

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

\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)

19 tháng 8 2021

b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)

\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)

\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)

\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)

\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)

\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)

NV
13 tháng 4 2019

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

\(K=\frac{sin\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}{cos\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}\)

\(K=\frac{sin\left(\frac{a+b}{2}\right)}{cos\left(\frac{a+b}{2}\right)}=tan\left(\frac{a+b}{2}\right)\)

23 tháng 4 2022

nhường cho I don't know :))

23 tháng 4 2022

Nó bt đếch gì mà nhường :) ?

6 tháng 5 2021

Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)

17 tháng 11 2019

vt lại đuề boài đi cậu, ko hịu nà :)