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a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
Mẫn Li
Câu 4 nếu bạn ko đánh sai thì người ghi đề sai :D, tử số phải là sinb chứ ko phải sina (đã chứng minh bên trên)
Câu 2b sửa lại thì cm dễ thôi:
\(\frac{cos\left(a+b\right).cos\left(a-b\right)}{sin^2a.sin^2b}=\frac{\frac{1}{2}cos2a+\frac{1}{2}cos2b}{sin^2a.sin^2b}=\frac{1-sin^2a-sin^2b}{sin^2a.sin^2b}=\frac{1}{sin^2a.sin^2b}-\frac{1}{sin^2a}-\frac{1}{sin^2b}\)
\(=\left(1+cot^2a\right)\left(1+cot^2b\right)-\left(1+cot^2a\right)-\left(1+cot^2b\right)\)
\(=1+cot^2a+cot^2b+cot^2a.cot^2b-2-cot^2a-cot^2b\)
\(=cot^2a.cot^2b-1\)
(từ đầu bằng thứ nhất ra thứ 2 sử dụng ct nhân đôi \(cos2x=1-2sin^2x\))
Rất xin lỗi bạn!
Câu 2b do mình đánh sai dấu phải là \(\frac{cos\left(a+b\right)\times cos\left(a-b\right)}{sin^2a\times sin^2b}=cot^2a\times cot^2b-1\)
Câu 3 mình cũng đánh sai luôn:
\(sin\frac{A}{2}=cos\frac{B}{2}\times cos\frac{C}{2}-sin\frac{C}{2}\times sin\frac{B}{2}\)
Còn câu 4 thì mình ko có đánh sai! Thành thật xin lỗi bạn! Mình sẽ khắc phục sự cố này!
\(A=\sqrt{\left(1-cos^2x\right)^2+4cos^2x}+\sqrt{\left(1-sin^2x\right)^2+4sin^2x}\)
\(=\sqrt{cos^4x+2cos^2x+1}+\sqrt{sin^4x+2sin^2x+1}\)
\(=\sqrt{\left(cos^2x+1\right)^2}+\sqrt{\left(sin^2x+1\right)^2}\)
\(=sin^2x+cos^2x+2=3\)
b/
\(3\left(sin^8x-cos^8x\right)=3\left(sin^4x+cos^4x\right)\left(sin^4x-cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)\)
\(=3sin^6x-3sin^4x.cos^2x+3sin^2x.cos^4x-3cos^6x\)
\(\Rightarrow B=-5sin^6x-3sin^4x.cos^2x+3sin^2x.cos^4x+cos^6x+6sin^4x\)
\(=-5sin^6x-3sin^4x\left(1-sin^2x\right)+3cos^4x\left(1-cos^2x\right)+cos^6x+6sin^4x\)
\(=-2sin^6x-2cos^6x+3sin^4x+3cos^4x\)
\(=-2\left(1-3sin^2x.cos^2x\right)+3\left(1-2sin^2x.cos^2x\right)\)
\(=-2+3=1\)
\(sinx+cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(=\sqrt{2}cos\left(\frac{\pi}{2}-\left(x+\frac{\pi}{4}\right)\right)=\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
\(sinx-cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx-\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
\(=-\sqrt{2}sin\left(\frac{\pi}{4}-x\right)=-\sqrt{2}cos\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x\)
\(=sin^2x-cos^2x+sin2x=sin2x-cos2x\)
\(=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)
Bạn ghi ko đúng đề
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(sina+sinb=2sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{2}}{4}\) (1)
\(cosa+cosb=2cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{6}}{2}\)
\(\Rightarrow cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{6}}{4}\) (2)
(1); (2) \(\Rightarrow tan\left(\frac{a+b}{2}\right)=\frac{\sqrt{3}}{3}\) \(\Rightarrow tan\left(a+b\right)=\sqrt{3}\) \(\Rightarrow a+b=60^0\)
\(\Rightarrow sin\left(a+b\right)=sin\left(60^0\right)=\frac{\sqrt{3}}{2}\)
\(sin100=sin\left(90+10\right)=cos10\)
\(sin\left(160\right)=sin\left(180^0-20^0\right)=sin20\)
\(cos170^0=cos\left(180^0-10^0\right)=-cos10^0\)
\(tan103^045'=tan\left(90^0+13^045'\right)=-cot13^045'\)
\(cot124^015'=cot\left(90^0+34^015'\right)=-tan34^015'\)
Tính tan15o
\(\cos^215^o=1-\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^2=\frac{8+2\sqrt{12}}{16}\)
\(=\frac{\left(\sqrt{6}\right)^2+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^2}{16}=\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{16}\)
Vì 15o<90o nên cos15o>0 => cos15o=\(\frac{\sqrt{6}+\sqrt{2}}{4}\)
tan 15o \(=\frac{sin15^0}{c\text{os}15^0}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)}{6-2}\right)^2=2-\sqrt{3}\)
CM:
\(2\sin15^0\cos15^0=2\frac{\sqrt{6}-\sqrt{2}}{4}.\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{1}{2}=\sin30^0\)