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a: sin a=2/3
=>cos^2a=1-(2/3)^2=5/9
=>\(cosa=\dfrac{\sqrt{5}}{3}\)
\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)
\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
b: cos a=1/5
=>sin^2a=1-(1/5)^2=24/25
=>\(sina=\dfrac{2\sqrt{6}}{5}\)
\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)
\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
c: cot a=1/tana=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>1/cos^2a=1+4=5
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
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
~ ~ ~ Áp dụng đẳng thức \(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\) ~ ~ ~
a)
\(\left(\sin\alpha+\cos\alpha\right)^2-2\sin\alpha\cos\alpha-1\)
\(=\left(\sin\alpha+\cos\alpha\right)^2-\left(2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha\right)\)
\(=\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha+\cos\alpha\right)^2\)
= 0
b)
\(\left(\sin\alpha-\cos\alpha\right)^2+2\sin\alpha\cos\alpha+1\)
\(=\left(\sin\alpha-\cos\alpha\right)^2+2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha\)
\(=\left(\sin\alpha-\cos\alpha\right)^2+\left(\sin\alpha+\cos\alpha\right)^2\)
\(=2\left(\sin^2\alpha+\cos^2\alpha\right)\)
= 2
c)
\(\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2+2\)
\(=2\left(\sin^2\alpha+\cos^2\alpha\right)+2\)
= 4
d)
\(\sin^2\alpha\cot^2\alpha+\cos^2\alpha\tan^2\alpha\)
\(=\left(\sin\times\dfrac{\cos}{\sin}\right)^2+\left(\cos\times\dfrac{\sin}{\cos}\right)^2\)
= 1
Bài 2:
\(1+\tan ^2a=1+\frac{\sin ^2a}{\cos ^2a}=\frac{\cos ^2a+\sin ^2a}{\cos ^2a}=\frac{1}{\cos ^2a}\)
\(1+\cot ^2a=1+\frac{\cos ^2a}{\sin ^2a}=\frac{\sin ^2a+\cos ^2a}{\sin ^2a}=\frac{1}{\sin ^2a}\)
Ta có đpcm.
1.
$0< a< 90^0\Rightarrow `1>\sin a, \cos a>0$
Do đó:
$\sin a-\tan a=\sin a-\frac{\sin a}{\cos a}=\frac{\sin a(\cos a-1)}{\cos a}<0$
$\Rightarrow \sin a< \tan a$
(đpcm)
$\cos a-\cot a=\cos a-\frac{\cos a}{\sin a}=\frac{\cos a(\sin a-1)}{\sin a}<0$
$\Rightarrow \cos a< \cot a$ (đpcm)