cho \(\sin a=\dfrac{3}{5}\) và \(\dfrac{\pi}{2}< a< \pi\)
tính cos a, cos 2a, cos\(\dfrac{a}{2}\)
cho \(\sin a=\dfrac{3}{5}\) và \(\dfrac{\pi}{2}< a< \pi\)
tính cos a, cos 2a, cos\(\dfrac{a}{2}\)
* Ta có: sin2a + cos2a = 1
⇒ cos a = √(1 - cos2a)
⇔ cos a = √(1 - 9/25)
Mà ta lại có π/2 < a< π ⇒ cos a ∈ [ 90o;180o ], cos a âm
⇒ cos a = -4/5
* Ta có: cos 2a = 1- 2sin2a
⇔ cos 2a = 1 - 2(9/25)
có π/2 < a< π ⇒ cos a âm
⇒ cos a = -7/25
trong mặt phẳng oxy cho tan giác ABC B(2;-3) C(3;-2) Viết pt đường tròn tâm I(1;-2) tiếp xúc với BC
\(\overrightarrow{BC}=\left(1;1\right)\)
=>VTPT là (-1;1)
PT của BC là:
\(-1\left(x-2\right)+1\left(y+3\right)=0\)
=>-x+2+y+3=0
=>-x+y+5=0
\(R=d\left(I;BC\right)=\dfrac{\left|1\cdot\left(-1\right)+\left(-2\right)\cdot1+5\right|}{\sqrt{1^2+1^2}}=\sqrt{2}\)
Phương trình đường tròn là:
\(\left(x-1\right)^2+\left(y+2\right)^2=2\)
Chứng minh các đẳng thức sau:
a, \(\sin^4\alpha-\cos^4\alpha+1=2\sin^2\alpha\)
b,\(\dfrac{\sin^2\alpha+2\cos^2\alpha-1}{\cot^2\alpha}=\sin^2\alpha\)
c, \(\dfrac{1-\sin^2\alpha.\cos^2\alpha}{\cos^2\alpha}-\cos^2\alpha=\tan^2\alpha\)
d, \(\dfrac{\sin^2\alpha-\tan^2\alpha}{\cos^2\alpha-\cot^2\alpha}=\tan^6\alpha\)
e, \(\left(1+\cot\alpha\right)\sin^3\alpha+\left(1+\tan\alpha\right)\cos^3\alpha=\sin\alpha.\cos\alpha\)
f,\(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-1}{\cot\alpha-\sin\alpha.\cos\alpha}=2\tan^2\alpha\)
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
e)
\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)
\(=(\sin ^3a+\cos ^3a)+\cot a.\sin ^3a+\tan a.\cos^3a\)
\(=(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)+\frac{\cos a}{\sin a}.\sin ^3a+\frac{\sin a}{\cos a}.\cos ^3a\)
\(=(\sin a+\cos a)(1-\sin a\cos a)+\cos a\sin ^2a+\sin a\cos ^2a\)
\(=\sin a+\cos a-\sin a\cos a(\sin a+\cos a)+\cos a\sin a(\sin a+\cos a)\)
\(=\sin a+\cos a\)
Nếu \(\sin x=3\cos x\)thì \(\sin x.\cos x\)bằng bao nhiêu
\(sin^2x+cos^2x=1\Leftrightarrow9cos^2x+cos^2x=1\Leftrightarrow cosx=\pm\dfrac{1}{\sqrt{10}}\Rightarrow sinx=\pm\dfrac{3}{\sqrt{10}}\Rightarrow sinxcosx=\dfrac{3}{10}\)
Giả sử A = tan \(x\) . tan (\(\dfrac{\pi}{3}-x\)) . tan \(\left(\dfrac{\pi}{3}+x\right)\) được rút gọn thành A = tan nx khi đó n bằng bao nhiêu
Rút gọn \(P=\sin\left(-\alpha\right)+\sin^2\left(\pi+\alpha\right)+\cos\left(\dfrac{\pi}{2}-\alpha\right)+\cos^2\left(\pi-\alpha\right)\)
Chứng minh rằng: sin 5x - 2 sin x(cos 4x + cos 2x) = sin x
\(VT=\sin5x-2\sin x\cdot\cos4x-2\sin x\cdot\cos2x\)
\(=\sin5x-\left(\sin5x-\sin3x\right)-\left(\sin3x-\sin x\right)\)
\(=\sin5x-\sin5x+\sin3x-\sin3x+\sin x\)
\(=\sin x=VP\)
Cho \(\Delta ABC\) cân chứng minh rằng:
\(a=2b.cosC\)
Chứng minh rằng:
tanx + cotx = 2/sinx
Rút gọn biểu thức : P=\(\dfrac{1+sin6x-cos6x}{1+sin6x+cos6x}\) sau đó tính P khi x= \(\dfrac{7\pi}{4}\)
\(P=\dfrac{1+2sin3xcos3x-\left(1-2sin^23x\right)}{1+2sin3xcos3x+2cos^2x-1}=\dfrac{2sin3xcos3x+2sin^23x}{2sin3xcos3x+2cos^23x}=\dfrac{sin3x}{cos3x}=tan3x\)
\(x=\dfrac{7\pi}{4}\Rightarrow P=tan\dfrac{21\pi}{4}=tan\dfrac{\pi}{4}=1\)