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NN
1
23 tháng 4 2017
Giao lưu
Đang tiếp cận đến sin cos
Lời giải
\(A=sin\left(2x\right).\left[2sin\left(3x\right).sin\left(x\right)\right]\)
\(A=sin\left(2x\right).\left\{-\left[cos\left(\dfrac{3x+x}{2}\right)-cos\left(\dfrac{3x-x}{2}\right)\right]\right\}\)
\(A=sin\left(2x\right).\left[cos\left(x\right)-cos\left(2x\right)\right]\)
\(A=sin\left(2x\right)cos\left(x\right)-sin\left(2x\right)cos\left(2x\right)\)
\(2A=\left[sin\left(\dfrac{2x+x}{2}\right)+sin\left(\dfrac{2x-x}{2}\right)\right]-\left[sin\left(\dfrac{2x+2x}{2}\right)+sin\left(\dfrac{2x-2x}{2}\right)\right]\)\(2A=sin\left(\dfrac{3x}{2}\right)+sin\left(\dfrac{x}{2}\right)-sin\left(2x\right)-0\)
\(A=\dfrac{sin\left(\dfrac{3x}{2}\right)+sin\left(\dfrac{x}{2}\right)-sin\left(2x\right)}{2}\)
MA
1
AH
Akai Haruma
Giáo viên
30 tháng 4 2021
Lời giải:
$A=\sin ^2a-\sin ^2b=(\sin a-\sin b)(\sin a+\sin b)$
$B$ không biến đổi được. Bạn coi lại đề.
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 4 2021
\(A=2sin\dfrac{a+b}{2}cos\dfrac{a-b}{2}+2sin\dfrac{a+b}{2}cos\dfrac{a+b}{2}\)
\(=2sin\dfrac{a+b}{2}\left(cos\dfrac{a+b}{2}+cos\dfrac{a-b}{2}\right)\)
\(=2sin\dfrac{a+b}{2}.2cos\dfrac{a}{2}cos\dfrac{b}{2}\)
\(=4sin\dfrac{a+b}{2}cos\dfrac{a}{2}cos\dfrac{b}{2}\)
TD
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
\(A=\dfrac{1}{2}cosa\left[cos\left(b+c\right)+cos\left(b-c\right)\right]\)
\(=\dfrac{1}{2}cosa.cos\left(b+c\right)+\dfrac{1}{2}cosa.cos\left(b-c\right)\)
\(=\dfrac{1}{4}cos\left(a+b+c\right)+\dfrac{1}{4}cos\left(a-b-c\right)+\dfrac{1}{4}cos\left(a+b-c\right)+\dfrac{1}{4}cos\left(a-b+c\right)\)
\(B=\dfrac{1}{2}sin2a\left(cos2a-cos10a\right)=\dfrac{1}{2}sin2a.cos2a-\dfrac{1}{2}sin2a.cos10a\)
\(=\dfrac{1}{4}sin4a-\dfrac{1}{4}sin12a+\dfrac{1}{4}sin8a\)
\(C=\dfrac{1}{2}\left(cos\dfrac{\pi}{3}-cos2x\right)cos2x=\dfrac{1}{2}cos2x\left(\dfrac{1}{2}-cos2x\right)\)
\(=\dfrac{1}{4}cos2x-\dfrac{1}{2}cos^22x=\dfrac{1}{4}cos2x-\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{2}cos4x\right)\)
\(=\dfrac{1}{4}cos2x-\dfrac{1}{4}cos4x-\dfrac{1}{4}\)
DH
1
13 tháng 8 2021
\(a)\)
\(1-sin\left(x\right)\)
\(=sin^2\frac{x}{2}+cos^2\frac{x}{2}-2.sin\frac{x}{2}.cos\frac{x}{2}\)
\(=\left(sin\frac{x}{2}-cos\frac{x}{2}\right)^2\)
\(b)\)
\(1+sin\left(x\right)\)
\(=sin^2\frac{x}{2}+cos^2\frac{x}{2}+2.sin\frac{x}{2}.cos\frac{x}{2}\)
\(=\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2\)
\(d)\)
\(1+2cos\left(x\right)\)
\(=2\left(\frac{1}{2}+cosx\right)\)
\(=2\left(cos60^o+cosx\right)\)
\(=4\left(cos\frac{60^o+x}{2}cos\frac{60^o-x}{2}\right)\)
\(=4cos\left(30^o+\frac{x}{2}\right)cos\left(30^o-\frac{x}{2}\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)
b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)
c/\(P=sin\left(30+60\right)=sin90=1\)
d/
\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)
\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)
e/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)