Lượng giác (chứng minh đẳng thức)

T

thanhphovinhyen

N

nguyenbahiep1

[laTEX] \\ \\ \frac{2tana+2tanb}{1-tana.tanb} = \frac{3sina}{cosa} \\ \\ \Leftrightarrow 2sina + 2tanb.cosa = 3sina- 3sina.tana.tanb \\ \\ \frac{2sinb.cosa}{cosb} = sina - \frac{3sin^2a.sinb}{cosa.cosb} \\ \\ 2sinb.cos^2a = sina.cosa.cosb - 3sin^2a.sinb \\ \\ 2sinb + sin^2a.sinb = \frac{1}{2}sin2a.cosb \\ \\ 4sinb + (1-cos2a).sinb = sin2a.cosb \\ \\sin2a.cosb + cos2a.sinb = 5sinb \\ \\ sin(2a+b) = 5sinb [/laTEX]
 
T

thanhphovinhyen

Lượng giác (tính giá trị biểu thức )

$\begin{array}{l}
1/\;{\mathop{\rm s}\nolimits} {\rm{in10}}^\circ .{\mathop{\rm s}\nolimits} {\rm{in50}}^\circ {\rm{.sin70}}^\circ \\
{\rm{2/}}\;{\rm{tan9}}^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ \\
3/\;\sin \frac{\pi }{{16}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} \\
4/\;\frac{{\tan 80^\circ }}{{\cot 25^\circ + \cot 75^\circ }} - \frac{{\cot 10^\circ }}{{\tan 25^\circ + \tan 75^\circ }} \\
5/\;\frac{{4 + 7\sin 2x}}{5}\;\;\;khi\;\tan x = 0,2 \\
6/\;\tan \left( {\frac{\pi }{4} - 2x} \right)\;\;\;khi\;\;\tan x = 2 \\
7/\;\;\sin 2x\;\;\;khi\;\;\cos x - \sin x\frac{1}{4} \\
8/\cos \frac{x}{2}\;\;khi\;\sin x = \frac{{ - 12}}{{13}}\;;\;\;\pi < x < \frac{{3\pi }}{2} \\
9/\;\sin \frac{a}{2}\;\;\;khi\;\;\sin a = 0,8\;;0 < x < \frac{\pi }{2} \\
\end{array}$
 
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L

levietdung1998

Câu 1

$A = \sin 10^\circ .\sin 30^\circ .\sin 70^\circ $

$\begin{array}{l}
A.c{\rm{os}}10^\circ = \frac{1}{2}\sin 20^\circ .c{\rm{os}}20^\circ .\sin 50^\circ = \frac{1}{4}.\sin 40^\circ .c{\rm{os}}40^\circ = \frac{1}{8}.\sin 80^\circ = \frac{1}{8}.c{\rm{os}}10^\circ \\
= > A = \frac{1}{8} \\
\end{array}$
 
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L

levietdung1998

Câu 3

\[\sin \frac{\pi }{{2.8}}.\cos \frac{\pi }{{2.8}}.\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{2}.\frac{1}{2}\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{8}\]
 
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X

xuanquynh97

$cot25^o+cot75^o=\dfrac{sin100^o}{sin25^o.sin75^o}$

\Rightarrow $\dfrac{tan80^o}{cot25^o+cot75^o}=\dfrac{sin25^o.sin75^o}{cos80^o}$

Tương tự

$\dfrac{cot10^o}{tan25^o+tan75^o}=\dfrac{cos25^o.cos75^o}{sin10^o}$

\Rightarrow $\dfrac{tan80^o}{cot25^o+cot75^o}-\dfrac{cot10^o}{tan25^o+tan75^o}=dfrac{sin25^o.sin75^o}{cos80^o}-\dfrac{cos25^o.cos75^o}{sin10^o}$


$=\dfrac{cos100^o}{cos80^o}$
 
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X

xuanquynh97

Bài 5: $sin^2x+cos^2=1$

\Rightarrow $sinx, cosx$

Tính được $sinxcosx$

Chú ý $tanx$ dương nên $sinx,cosx$ cùng dấu

Bài 6 Tương tự bài 5

$tan(\dfrac{\pi}{4}-2x)=\dfrac{sin2x-cos2x}{sin2x+cos2x}$
 
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