2)
a) Cho cos α = \(\dfrac{1}{3}\). Tính giá trị P = 3.sin2 α + 4.cos2 α .
b) Cho tan α = \(\dfrac{3}{4}\). Tính sin α ; cos α ; cot α .
c) Cho tan α = \(\dfrac{1}{2}\). Tính \(\dfrac{cosa-sina}{cosa+sina}\) ( α nhọn ).
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Những câu hỏi liên quan
26 tháng 7 2023
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: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
\(A=\dfrac{2tan^2a+\dfrac{5}{cos^2a}}{4-\dfrac{3}{cos^2a}}=\dfrac{2tan^2a+5\left(1+tan^2a\right)}{4-3\left(1+tan^2a\right)}=...\) (bạn tự thay số bấm máy nhé)
\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)
9 tháng 8 2023
1:
a: sin a=căn 3/2
\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)
\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)
cot a=1/tan a=1/căn 3
b: \(tana=2\)
=>cot a=1/tan a=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=5\)
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)
c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
tan a=5/13:12/13=5/12
cot a=1:5/12=12/5
21 tháng 3 2022
\(sin\left(\text{α}-\dfrac{\Pi}{4}\right)-cos\left(\text{α}-\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.cos\dfrac{\Pi}{4}-cos\text{α}-sin\dfrac{\Pi}{4}-\left(cos\text{α}.cos\dfrac{\Pi}{4}+sin\text{α}.sin\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-sin\text{α}.\dfrac{\sqrt{2}}{2}\)
\(=\dfrac{-2\sqrt{2}}{6}\)
\(=\dfrac{-\sqrt{2}}{3}\)
HP
19 tháng 8 2021
a, \(\dfrac{1-sin2a}{1+sin2a}\)
\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)
\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)
\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)
HP
19 tháng 8 2021
b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)
\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)
\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)
\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)
\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)
\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)
LH
2
28 tháng 6 2021
\(sin\alpha^2+cos\alpha^2=1\Rightarrow sin\alpha^2=1-cos\alpha^2=1-\dfrac{1}{25}=\dfrac{24}{25}\Rightarrow sin\alpha=\dfrac{2\sqrt{6}}{5}\)
\(\Rightarrow cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{1}{5}:\dfrac{2\sqrt{6}}{5}=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{24}\)
28 tháng 6 2021
\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\sin^2\alpha=1-\dfrac{1}{25}=\dfrac{24}{25}\)
hay \(\sin\alpha=\dfrac{2\sqrt{6}}{5}\)
\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)
\(\cot\alpha=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
28 tháng 6 2021
b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)
hay \(\cos\alpha=\dfrac{4}{5}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)
\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)
\(=\dfrac{141}{25}\)
c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)
\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)
2 tháng 10 2021
a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)
\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)
2 tháng 10 2021
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
15 tháng 7 2021
a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha\)
\(=5+\dfrac{16}{25}=\dfrac{141}{25}\)
Lời giải:
a) Áp dụng công thức \(\sin ^2a+\cos ^2a=1\) thì:
\(P=3\sin ^2a+4\cos ^2a=3(\sin ^2a+\cos ^2a)+\cos ^2a\)
\(=3.1+(\frac{1}{3})^2=\frac{28}{9}\)
b)
\(\tan a=\frac{3}{4}\Rightarrow \cot a=\frac{1}{\tan a}=\frac{4}{3}\)
\(\frac{3}{4}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{3}{4}\cos a\)
\(\Rightarrow \sin ^2a=\frac{9}{16}\cos ^2a\)
\(\Rightarrow \sin ^2a+\cos ^2a=\frac{25}{16}\cos ^2a\Rightarrow \frac{25}{16}\cos ^2a=1\)
\(\Rightarrow \cos ^2a=\frac{16}{25}\Rightarrow \cos a=\pm \frac{4}{5}\)
Nếu \(\Rightarrow \sin a=\pm \frac{3}{5}\) (theo thứ tự)
c)
\(\frac{1}{2}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{\cos a}{2}\). Vì a góc nhọn nên \(\cos a\neq 0\)
Do đó:
\(\frac{\cos a-\sin a}{\cos a+\sin a}=\frac{\cos a-\frac{\cos a}{2}}{\cos a+\frac{\cos a}{2}}=\frac{\cos a(1-\frac{1}{2})}{\cos a(1+\frac{1}{2})}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}\)