Tính giá trị của các bt sau(ko dùng máy tính) b/ B = 3cos 10 ° - 4cos^3 (10 °) . c/C = sin 30 ° * (2 - 4cos^2 15 °) d / D = 4sin^3 (40 °) + 3cos 130 °
Tính giá trị C = 3 sin α + 4 cos α 2 + 4 sin α - 3 cos α 2
A. 25
B. 16
C. 9
D. 25 + 48sin α .cos α
1) Cho \(3\sin^4x-cos^4x=\frac{1}{2}\), tìm giá trị của B=\(sin^4x+3cos^4x\)
2) Cho \(4sin^4x+3cos^4x=\frac{7}{4}\), tìm giá trị của C=\(3sin^4x+4cos^4x\)
\(3sin^4x-\left(1-sin^2x\right)^2=\frac{1}{2}\Leftrightarrow3sin^4x-\left(sin^4x-2sin^2x+1\right)=\frac{1}{2}\)
\(\Leftrightarrow2sin^4x+2sin^2x-\frac{3}{2}=0\) \(\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\\sin^2x=-\frac{3}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cos^2x=1-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow B=\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)^2=1\)
\(4sin^4x+3\left(1-sin^2x\right)^2=\frac{7}{4}\Leftrightarrow4sin^4x+3\left(sin^4x-2sin^2x+1\right)=\frac{7}{4}\)
\(\Leftrightarrow7sin^4x-6sin^2x+\frac{5}{4}=0\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\Rightarrow cos^2x=\frac{1}{2}\\sin^2x=\frac{5}{14}\Rightarrow cos^2x=\frac{9}{14}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}C=3\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)^2=\frac{7}{4}\\C=3\left(\frac{5}{14}\right)^2+4\left(\frac{9}{14}\right)^2=\frac{57}{28}\end{matrix}\right.\)
cho tanθ=3 tính giá trị:
A = \(\dfrac{2sin\theta-3cos\theta}{4cos\theta+5sin\theta}\) B = \(\dfrac{sin^2\theta-4sin\theta cos\theta+3cos^2\theta}{5-2sin^2\theta}\)
\(tanx=3\) \(\Leftrightarrow sinx=3cosx\)
\(A=\dfrac{2.3.cosx-3cosx}{4cosx+5.3cosx}=\dfrac{3cosx}{19cosx}=\dfrac{3}{19}\)
\(B=\dfrac{sin^2x-4sinxcosx+3cos^2x}{5-2sin^2x}\)
\(=\dfrac{\left(3cosx\right)^2-4.3cosx.cosx+3cos^2x}{5-2\left(3cosx\right)^2}\)
\(=\dfrac{9cos^2x-12cos^2x+3cos^2x}{5-18cos^2x}=0\)
giải các phường trình sau:
a/\(sin^3x+cos^3x=sinx+cosx\)
b/\(sin^3x+2sin^2xcosx-3cos^3x=0\)
c/\(3cos^4x-4cos^2xsin^2x-sin^4x=0\)
d/\(sinx-4sin^3x+cosx=0\)
mọi người giúp em với em cảm ơn mọi người nhìu
\(a\text{) }sin^3x+cos^3x=sinx+cosx\\ \Leftrightarrow\left(sinx+cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)=sinx+cosx\\ \Leftrightarrow-\frac{1}{2}sin2x\left(sinx+cosx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=-cosx=sin\left(x-\frac{\pi}{2}\right)\\sin2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{2}-x+a2\pi\\2x=b\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{4}+a\pi\\x=\frac{b\pi}{2}\end{matrix}\right.\)
\(\text{b) }sin^3x+2sin^2x\cdot cosx-3cos^3x=0\\ \Leftrightarrow\left(sin^3x-cos^3x\right)+2cosx\cdot\left(sin^2x-cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(sinx\cdot cosx+1\right)+\left(sinx-cosx\right)\left(2sinx\cdot cosx+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(3sinx\cdot cosx+1+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(\frac{3}{2}sin2x+2+cos2x\right)=0\)
Với \(sinx-cosx=0\)
\(\Leftrightarrow sinx=cosx=sin\left(\frac{\pi}{2}-x\right)\\ \Leftrightarrow x=\frac{\pi}{2}-x+a2\pi\\ \Leftrightarrow x=\frac{\pi}{4}+a\pi\)
Với \(\frac{3}{2}sin2x+2+cos2x=0\)
\(\Leftrightarrow sin^22x+\left(\frac{3}{2}sin2x+2\right)^2=1\left(VN\right)\)
\(\text{c) }3cos^4x-4cos^2x\cdot sin^2x-sin^4x=0\)
Nhận thấy sinx=0 không là nghiệm pt.
Chia cả 2 vế cho sin4x ta được
\(pt\Leftrightarrow\frac{3cos^4x}{sin^4x}-\frac{4cos^2x}{sin^2x}-1=0\\ \Leftrightarrow3cot^4x-4cot^2x-1=0\\ \Leftrightarrow cot^2x=\frac{2+\sqrt{7}}{3}\\ \Leftrightarrow cotx=\pm\sqrt{\frac{2+\sqrt{7}}{3}}\\ \Leftrightarrow x=arccot\left(\pm\sqrt{\frac{2+\sqrt{7}}{3}}\right)+k2\pi\)
d) kiểm tra đề.
giai pt:
a) \(4sin^5x.cosx-4cos^5x.sinx=sin^24x\)
b) \(4sin^2\frac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\frac{3\pi}{4}\right)\)
c) \(sin^2\left(x+\frac{\pi}{3}\right)+sinx+\sqrt{3}cosx=\frac{5}{4}\)
d) \(2sinx\left(1+cos2x\right)+sin2x=1+2cosx\)
e) \(sin^2x+4sinx.cosx+3cos^2x-sinx-3ccosx=0\)
a/
\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)
\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)
\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)
\(\Leftrightarrow-sin4x=sin^24x\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=k\pi\\4x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
b/
\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)
\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)
\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
Cho góc nhọn α , tính giá trị của biểu thức
A=(3sinα+4cosα)2 + ( 4sinα-3cosα)2
Ta dễ chứng minh \(sin^2a+cos^2a=1\) theo định lí Pytago
\(A=\left(3sina+4cosa\right)^2+\left(4sina-3cosa\right)^2\)
\(A=9sin^2a+24sina.cosa+16cos^2a+16sin^2a-24sina.cosa+9cos^2a\)
\(A=25sin^2a+25cos^2a=25\)
8cos^3(x)-3cos^2(x)-4cos(x)+sin^2(x)=0
giúp em câu này ạ!
\(\Leftrightarrow8\cos^3x-3\cos^2x-4\cos x+\left(1-\cos^2x\right)=0\)
\(\Leftrightarrow8\cos^3x-4\cos^2x-4\cos x+1=0\)
Sau đó ấn máy tính nhưng bài này ra nghiệm lẻ lắm bạn
Không dùng máy tính bỏ túi. Hãy tính 3cos2a-4sin2a, biết sin a =0,2
sin a = 0,2
=> sin2 a = 0,4
cos 2 a = 0,6
Thế vô được 0,2
Sin a = 0,2
=> sin2 a = 0,04
=> cos 2 a = 0,96
=> 3cos2a-4sin2a = 2,72
Bài 1 CM các đẳng thức sau:
a, 1+ sin2a / sina + cosa - 1-tan ²a/2 / 1+ tan ²a/2 = sina
b, cota - tana = 2cot2a
c, 1+ cosa +cos2a + cos3a/ 2cos²a + cosa-1 = 2cosa
d, sin²a / sina- cosa - sina + cosa / tan²a = sina + cosa
e, sin²a - cos²(a-b ) + 2coscosb ×cos(a-b) = cos2a
f, cos²a - 2sina × ( 1-sina ) × cosa +( 1 + sina) × cosa - 2×(1+sina ) / 1- sina = cosa
Bài 2 CM các đẳng thức sau ko phụ thuộc vào x
a, A= sin⁶x + cos⁶x - 1 / sin⁴x + cos ⁴x -1
b, B = ( 2sin ⁶x - 3sin ⁴x - 4sin²x ) +( 2cos⁶x - 3 cos⁴x- 4cos⁴x
c, C= sin⁴x + 3cos⁴x -1 / sin⁶x + cos⁶x + 3cos⁴x-1
Giải giúp tớ 2 bài này vs tớ cảm ơn nhìu