cho góc nhọn α :
chứng minh rằng: \(\frac{1-\tan\text{α}}{1+\tan\text{α}}\)=\(\frac{\cos\text{α}-\sin\text{α}}{\cos\text{α}+\sin\text{α}}\)
Chứng minh : \(\dfrac{sin^2\text{α}}{cos\text{α}\left(1+tan\text{α}\right)}-\dfrac{cos^2\text{α}}{sin\text{α}\left(1+cot\text{α}\right)}-sin\text{α}-cos\text{α}\)
Chứng minh R
sin4α + sin2α . cos2α + cos2α = 1
\(\dfrac{sin\text{α}}{1-cos\text{α}}\)+\(\dfrac{sin\text{α}}{1+cos\text{α}}\)+\(\dfrac{2}{sin\text{α}}\)
\(\dfrac{sin\text{α}}{1+cos\text{α}}\)+\(\dfrac{1+cos\text{α}}{sin\text{α}}\)=\(\dfrac{2}{sin\text{α}}\)
a: VT=sin^2a(sin^2a+cos^2a)+cos^2a
=sin^2a+cos^2a
=1=VP
b: \(VT=\dfrac{sina+sina\cdot cosa+sina-sina\cdot cosa}{1-cos^2a}=\dfrac{2sina}{sin^2a}=\dfrac{2}{sina}=VP\)
c: \(VT=\dfrac{sin^2a+1+2cosa+cos^2a}{sina\left(1+cosa\right)}\)
\(=\dfrac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\dfrac{2}{sina}=VP\)
Chứng minh các hệ thức:
a) \(\dfrac{cos\text{ α }}{1-sin\text{ α}}=\dfrac{1+sin\text{ α}}{cos\text{ α}}\)
b)\(\dfrac{\left(sin\text{ α }+cos\text{ α }\right)^2-\left(sin\text{ α }-cos\text{ α }\right)^2}{sin\text{ α }cos\text{ α }}=4\)
a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)
\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)
b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)
\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)
=4
Cho góc α
thỏa mãn `π\2`<α<π,cosα=−\(\dfrac{1}{\sqrt{3}}\). Tính giá trị của các biểu thức sau:
a) sin(α+\(\dfrac{\text{π}}{6}\))
b) cos(α+$\frac{\text{π}}{6}$)
c) sin(α−$\frac{\text{π}}{3}$)
d) cos(α−$\frac{\text{π}}{6}$)
a: pi/2<a<pi
=>sin a>0
\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)
\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)
\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)
b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
c: \(sin\left(a-\dfrac{pi}{3}\right)\)
\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)
\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)
d: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
CMR: α<45* ta có công thức:
a/ \(sin^2\alpha=\frac{1-cos2\text{α}}{2}\)
b/ \(cos^2\text{α}=\frac{1+cos2\text{α}}{2}\)
c/ \(cos2\text{α}=cos^2\text{α}-sin^2\text{α}\)
Bài 1: Tính gt biểu thức: \(cos^220^o+cos^240^o+cos^250^o+cos^270^o\)
Bài 2:Chứng minh hệ thức:
a,\(cot^2\text{α}-cos^2\text{α}=cot^2\text{α}.cos^2\text{α}\)
b,\(\dfrac{1+cos\text{ α}}{sin\text{ α}}=\dfrac{sin\text{ α}}{1-cos\text{ α}}\)
(P/s: tại mik ko tìm đc kí hiệu Anpha nên phải viết chữ =.=)
Các bạn giúp mik vs, mik đang cần gấp ak.Mik cảm ơn!!!!
bài 1: ta có : \(cos^220+cos^240+cos^250+cos^270\)
\(=cos^220+cos^270+cos^240+cos^250\)
\(=cos^220+cos^2\left(90-20\right)+cos^240+cos^2\left(90-40\right)\)
\(=cos^220+sin^220+cos^240+sin^240=1+1=2\)
bài 2: a) ta có : \(cot^2\alpha-cos^2\alpha=cos^2\alpha\left(\dfrac{1}{sin^2\alpha}-1\right)=cos^2\alpha.\left(\dfrac{1-sin^2\alpha}{sin^2\alpha}\right)\)
\(=cos^2\alpha.\left(\dfrac{cos^2\alpha}{sin^2\alpha}\right)=cos^2\alpha.cot^2\alpha\left(đpcm\right)\)
b) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Leftrightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Leftrightarrow\dfrac{1+cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1-cos\alpha}\left(đpcm\right)\)
Tính:\(\frac{3\sin\text{α}+2cos\text{α}}{3sin\text{α}-2cos\text{α}}\) Biết tanα=8
\(A=\frac{\frac{3sina}{cosa}+\frac{2cosa}{cosa}}{\frac{3sina}{cosa}-\frac{2cosa}{cosa}}=\frac{3tana+2}{3tana-2}=\frac{24+2}{24-2}=\frac{26}{22}=\frac{13}{11}\)
cho sin α = \(\frac{2}{3}\)
Tính P= \(\frac{tan\text{α}-cos\text{α}}{cotg\text{α }}\)
Giải nhanh hộ mình
\(\sin\alpha=\frac{2}{3}\) nên a là góc nhọn trong tam giác vuông có cạnh đối là 2, cạnh huyền là 3 suy ra cạnh kề = \(\sqrt{5}\)
Vậy: \(\cos\alpha=\sqrt{\frac{5}{3}};\tan\alpha=\frac{2}{\sqrt{5}};\cot\alpha=\sqrt{\frac{5}{2}}\)
Cho tan α = 2 . Tính \(\frac{2cos\text{α + sinα}}{sin\text{α - 3 cosα }}\)
Lên cốc cốc tìm cốc cốc toán thay 2 vào mà tìm vậy cũng phải đăng
Thay 2 vào chỗ nào có a xong rồi lên cốc côc viết ý hệt ra nhưng thay mỗi a =2