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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/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
Tính giá trị của biểu thức :
a) \(2\sin30^0+3\cos45^0-\sin60^0\)
b) \(2\cos30^0+3\sin45^0-\cos60^0\)
a)
\(2sin30+3sin45^o-sin60^o=2.\dfrac{1}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}\)\(=\dfrac{2+3\sqrt{2}-\sqrt{3}}{2}\).
b)\(2cos30^o+3sin45^o-cos60^o=2.\dfrac{\sqrt{3}}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)\(=\dfrac{2\sqrt{3}+3\sqrt{2}-1}{2}\).
Ta có sin0 độ + cos0 độ =0+1=1 nên A sai.
sin90 độ + cos90 độ =1+0=1 nên B đúng.
sin180 độ + cos180 độ=0+(−1)=−1 nên C đúng.
sin60 độ + cos60 độ =√3/2+1/2=√3+1/2 nên D đúng.
Chọn (A).
Lời giải:
Đặt $a-\frac{b}{2}=x; \frac{a}{2}-b=y$ thì $45^0< x< 180^0; -45^0< y< 90^0$
$\cos x=\frac{-1}{4}; 45^0< x< 180^0$ nên $\sin x=\frac{\sqrt{15}}{4}$
$\sin y=\frac{1}{3}; -45^0< y< 90^0$ nên $\cos y=\frac{2\sqrt{2}}{3}$
\(P=72\cos (2x-2y)+49=72[2\cos ^2(x-y)-1]+49=144\cos ^2(x-y)-23\)
\(=144(\cos x\cos y+\sin x\sin y)^2-23=-4\sqrt{30}\)
Đáp án C.
\(cos\left[\frac{\pi}{3}+\left(2k+1\right)\pi\right]=cos\left(\frac{\pi}{3}+\pi+2k\pi\right)=cos\left(\frac{\pi}{3}+\pi\right)=-cos\frac{\pi}{3}=-\frac{1}{2}\)
a3+b3+c3=3abc
<=>(a+b)3-3ab(a+b)-3abc+c3=0
<=>(a+b+c)[(a+b)2-(a+b)c+c2]-3ab.(a+b+c)=0
<=>(a+b+c)(a2+b2+c2-ab-bc-ac)=0
<=>(a+b+c)(2a2+2b2+2c2-2ab-2bc-2ac)=0
<=>(a+b+c)[(a-b)2+(b-c)2+(c-a)2]=0
<=>a+b+c=0 [(a-b)2+(b-c)2+(c-a)2 khác 0]
=>a2+b2-c2=-2ab;b2+c2-a2=-2bc;c2+a2-b2=-2ac
Suy ra : P=\(-\left(\dfrac{1}{2ab}+\dfrac{1}{2bc}+\dfrac{1}{2ac}\right)=-\dfrac{a+b+c}{2abc}=0\)
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$