Chương 6: CUNG VÀ GÓC LƯỢNG GIÁC. CÔNG THỨC LƯỢNG GIÁC - Hỏi đáp

Hình vẽ:

Lời giải:

Gọi $I(-1;3)$ là tâm của đường tròn. Kẻ $IH\perp (\Delta)$ thì $H$ chính là trung điểm của $AB$.

Ta có:

\(IH=d(I,\Delta)=\frac{|-1-2.3+5|}{\sqrt{1^2+(-2)^2}}=\frac{2\sqrt{5}}{5}\)

\(IA=R=\sqrt{20}\)

Áp dụng định lý Pitago: \(AH=\sqrt{IA^2-IH^2}=\sqrt{20-(\frac{2\sqrt{5}}{5})^2}=\frac{4\sqrt{30}}{5}\)

\(\Rightarrow AB=2AH=\frac{8\sqrt{30}}{5}\)

e)

\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)

\(=(\sin ^3a+\cos ^3a)+\cot a.\sin ^3a+\tan a.\cos^3a\)

\(=(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)+\frac{\cos a}{\sin a}.\sin ^3a+\frac{\sin a}{\cos a}.\cos ^3a\)

\(=(\sin a+\cos a)(1-\sin a\cos a)+\cos a\sin ^2a+\sin a\cos ^2a\)

\(=\sin a+\cos a-\sin a\cos a(\sin a+\cos a)+\cos a\sin a(\sin a+\cos a)\)

\(=\sin a+\cos a\)

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)

\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)

\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)

\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)

Lời giải:
\(\sqrt{(1-\tan ^2x)(\cot ^2x-1)}=\sqrt{(1-\tan ^2x)[\cot^2x-(\tan x\cot x)^2]}\)

\(=\sqrt{(1-\tan ^2x)\cot ^2x(1-\tan ^2x)}=\sqrt{\cot ^2x(1-\tan ^2x)^2}\)

\(=|\cot x(1-\tan ^2 x)|=|\cot x-\cot x.\tan x\tan x|=|\cot x-\tan x|\)

Mấu chốt trong các bài này là việc sử dụng công thức $\sin ^2a+\cos ^2a=1$

Lời giải:

a)

\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)

\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)

b)

\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)

\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)

c)

\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)

d)

\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)

\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)

\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)

a) Sin (B+C) = Sin (180-A) = Sin A
b) Cos (A+B) = Cos ( 180-A) = Cos A
c) Sin (\(\dfrac{B+C}{2}\)) = Sin \(\left(\dfrac{180-A}{2}\right)\)= Sin \(\left(90^0-\dfrac{A}{2}\right)\)= Cos \(\dfrac{A}{2}\)

d) Tan \(\left(\dfrac{A+C}{2}\right)\)= Tan\(\left(\dfrac{180-B}{2}\right)\)=Tan\(\left(90^0-\dfrac{B}{2}\right)\)= Cot \(\dfrac{B}{2}\)

\(A=cos^21+coss^22+...+cos^288+cos^289-\frac{1}{2}\)

\(A=1-sin^21+1-sin^22+...+1-sin^244+cos^245+cos^246+...+cos^289-\frac{1}{2}\)

\(A=1\cdot44+cos^245-\frac{1}{2}\)

\(A=44\)

B=\(sin^21+sin^22+...+sin^289-\frac{1}{2}\)

\(B=1-cos^21+1-cos^22+...+sin^245+sin^246+....+sin^289-\frac{1}{2}\)

\(B=1\cdot44+sin^245-\frac{1}{2}=44\)

\(C=tan^21\cdot tan^22\cdot...\cdot tan^288+tan^289\)

\(C=tan^21\cdot\left(tan^22\cdot tan^288\right)\cdot...\cdot\left(tan^244\cdot tan^246\right)\cdot tan^245+tan^289\)

\(C=tan^21+tan^289\approx3282\)

D = \(\left(tan^21:cot^289\right)+...+\left(tan^244:tan^246\right)+tan^245\)

\(D=\left(tan^21\cdot tan^289\right)+...+\left(tan^244\cdot tan^246\right)+tan^245\)

\(D=1+...+1+1\)

ta thấy từ 1 đến 89 có 89 số hạng, trong đó có 44 cặp.

vậy D = 45

\(G=\frac{cos2x-sin4x-cos6x}{cos2x+sin4x-cos6x}=\frac{-2sin4xsin2x-sin4x}{-2sin4xsin2x+sin4x}\)

\(G=\frac{-sin4x\left(2sin2x+1\right)}{-sin4x\left(2sin2x-1\right)}=\frac{2sin2x+1}{2sin2x-1}\)

\(F=\frac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)

\(F=\frac{2sin4xcos3x+sin4x}{2cos4xcos3x+cos4x}\)

\(F=\frac{2sin4x\left(cos3x+1\right)}{2cos4x\left(cos3x+1\right)}=tan4x\)