chứng minh đẳng thức $\frac{3 – 4cos2a + cos4a}{3 + 4cos2a + cos4a}$ = $tan^{4}$a 21/08/2021 Bởi Piper chứng minh đẳng thức $\frac{3 – 4cos2a + cos4a}{3 + 4cos2a + cos4a}$ = $tan^{4}$a
Đáp án: Giải thích các bước giải: Ta có: `VT= {3-4cos2a+cos4a}/{3+4cos2a+cos4a}` `={3-4cos2a+(2cos^2 2a-1)}/{3+4cos2a+(2cos^2 2a-1)}` `={2(cos^2 2a-2cos 2a +1)}/{2(cos^2 2a+2cos2a +1)}` `={(cos2a-1)^2}/{(cos2a+1)^2}` `={(1-2sin^2 a-1)^2}/{(2cos^2 a-1+1)^2}` `={4sin^4 a}/{4cos^4 a}` `=({sina}/{cosa})^4=tan^4 a=VP` Vậy `{3-4cos2a+cos4a}/{3+4cos2a+cos4a}=tan^4a` Bình luận
$\frac{3-4\cos 2a+\cos 4a}{3+4\cos 2a+\cos 4a}=\frac{3-4\cos \:2a+2\cos ^2\:2a-1}{3+4\cos \:2a+2\cos ^2\:2a-1}=\frac{2\cos \:^2\:2a-4cos2a+2}{2\cos \:\:^2\:2a+4cos2a+2}=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}=\left(\frac{cos2a-1}{cos2a+1}\right)^2=\left(\frac{-2sin^22a}{2cos^22a}\right)^2=tan^4a$ XIN CTLHN :((( 1 CÁI NỮA LÊN RANK Bình luận
Đáp án:
Giải thích các bước giải:
Ta có:
`VT= {3-4cos2a+cos4a}/{3+4cos2a+cos4a}`
`={3-4cos2a+(2cos^2 2a-1)}/{3+4cos2a+(2cos^2 2a-1)}`
`={2(cos^2 2a-2cos 2a +1)}/{2(cos^2 2a+2cos2a +1)}`
`={(cos2a-1)^2}/{(cos2a+1)^2}`
`={(1-2sin^2 a-1)^2}/{(2cos^2 a-1+1)^2}`
`={4sin^4 a}/{4cos^4 a}`
`=({sina}/{cosa})^4=tan^4 a=VP`
Vậy `{3-4cos2a+cos4a}/{3+4cos2a+cos4a}=tan^4a`
$\frac{3-4\cos 2a+\cos 4a}{3+4\cos 2a+\cos 4a}=\frac{3-4\cos \:2a+2\cos ^2\:2a-1}{3+4\cos \:2a+2\cos ^2\:2a-1}=\frac{2\cos \:^2\:2a-4cos2a+2}{2\cos \:\:^2\:2a+4cos2a+2}=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}=\left(\frac{cos2a-1}{cos2a+1}\right)^2=\left(\frac{-2sin^22a}{2cos^22a}\right)^2=tan^4a$ XIN CTLHN :((( 1 CÁI NỮA LÊN RANK