A= \(\frac{1}{2}\)[sin(-10)+sin90] +\(\frac{1}{2}\)(sin10+sin90)

A= \(\frac{1}{2}\)(-sin10 +1) +\(\frac{1}{2}\)(sin10 +1)

A=\(\frac{1}{2}\)(-sin10+sin10)+1

A= 1

Câu 5. Cho x,y dương thỏa mãn \(x+y=\dfrac{1}{2}\).Tìm giá trị nhỏ nhất của 

\(P=\dfrac{1}{x}+\dfrac{1}{y}\)

Giải:

\(P=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{\dfrac{1}{2}}{xy}=\dfrac{2}{xy}\)

--> P nhỏ nhất khi \(xy\) lớn nhất

Ta có:

\(x^2+y^2\ge2xy\) ( BĐT AM-GM )

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow1\ge4xy\)

\(\Leftrightarrow xy\le\dfrac{1}{4}\)

\(\Rightarrow P\ge2:\dfrac{1}{4}=8\)

Vậy \(Min_P=8\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{4}\)

a: \(\sin25^0< \sin70^0\)

b: \(\cos40^0>\cos75^0\)

c: \(\sin38^0=\cos52^0< \cos27^0\)

d: \(\sin50^0=\cos40^0>\cos50^0\)

Chú ý rằng: sin45⁰ = cos45⁰, sin40⁰ = cos50⁰, sin50⁰ = cos40⁰

Ta được:

\(\dfrac{\cos50^0-\cos45^0+\cos50^0}{\cos40^0-\cos45^0+\cos50^0}-\dfrac{6\times3\left(\dfrac{\sqrt{3}}{3}+\tan15^0\right)}{3\left(1-\dfrac{\sqrt{3}}{3}\tan15^0\right)}\)

\(=1-6\left(\dfrac{\tan30^0+\tan15^0}{1-\tan30^0\times\tan15^0}\right)\)

\(=1-6\tan45^0=-5\)

a/ \(\tan40.\cot40+\frac{\sin50}{\cos40}\)

\(=1+\frac{\cos40}{\cos40}=1+1=2\)

Đề yêu cầu làm gì bạn

thực hiện phép tính nha

Câu 3:

\(A=cos\frac{\pi}{7}.cos\frac{5\pi}{7}.cos\frac{4\pi}{7}=cos\frac{\pi}{7}.cos\left(\pi-\frac{2\pi}{7}\right).cos\frac{4\pi}{7}\)

\(A=-cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.2sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}\)

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

\(\Rightarrow A=\frac{1}{8}\)

Câu 4:

Đầu tiên ta chứng minh công thức:

\(tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}\)

Áp dụng để biến đổi tử số:

\(tan30+tan60+tan40+tan50=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{cos30.cos60}+\frac{1}{cos40.cos50}\)

\(=\frac{2}{cos90+cos30}+\frac{2}{cos90+cos10}=\frac{2}{cos30}+\frac{2}{cos10}=2\left(\frac{cos30+cos10}{cos30.cos10}\right)\)

\(=2\left(\frac{2cos20.cos10}{cos30.cos10}\right)=\frac{4.cos20}{cos30}=\frac{8\sqrt{3}}{3}.cos20\)

\(\Rightarrow A=\frac{\frac{8\sqrt{3}}{3}cos20}{cos20}=\frac{8\sqrt{3}}{3}\)

Câu 5:

\(cos54.cos4-cos36.cos86=cos54.cos4-cos\left(90-54\right).cos\left(90-4\right)\)

\(=cos54.cos4-sin54.sin4=cos\left(54+4\right)=cos58\)

Câu 1:

\(A=\frac{1}{2sin10}-2sin70=\frac{1-4sin10.sin70}{2sin10}=\frac{1+2\left(cos80-cos60\right)}{2sin10}\)

\(=\frac{1+2cos80-1}{2sin10}=\frac{2cos80}{2sin10}=\frac{sin10}{sin10}=1\)

Câu 2:

\(cos10.cos30.cos50.cos70=cos10.cos30.\frac{1}{2}\left(cos120+cos20\right)\)

\(=\frac{1}{2}cos30\left(cos10.cos120+cos10.cos20\right)\)

\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}\left(cos30+cos10\right)\right)\)

\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}cos30+\frac{1}{2}cos10\right)\)

\(=\frac{1}{2}.\frac{\sqrt{3}}{2}\left(-\frac{1}{2}cos10+\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}cos10\right)\)

\(=\frac{3}{16}\)

\(\cot65^0=\tan25^0< \cot60^0=\tan30^0< \tan50^0< \tan70^0\)

\(cos\left(2x-18^o\right).tan50^0+sin\left(2x-18^o\right)=\dfrac{1}{2cos130^0}\)

⇔\(cos\left(2x-18^o\right).sin50^0+sin\left(2x-18^o\right).cos50^0=\dfrac{cos50^0}{2cos130^0}\)

(Nhân cả 2 vế với cos50⁰)

⇔ sin (50⁰ + 2x - 18⁰) = \(-\dfrac{1}{2}\)

⇔ \(\left[{}\begin{matrix}2x+32^0=-30^0+k.360^0\\2x+32^0=210^0+k.360^0\end{matrix}\right.\) với k là số nguyên