tìm giá trị biểu thức Q=sin^2a+sin(π/6+a)sin(π/6-a)+cos(2a+5 π) biết cos(a)=1/4 20/08/2021 Bởi Rylee tìm giá trị biểu thức Q=sin^2a+sin(π/6+a)sin(π/6-a)+cos(2a+5 π) biết cos(a)=1/4
Đáp án: `Q=9/8` Giải thích các bước giải: Ta có: `\qquad cosa=1/4` `=>cos2a=2cos^2a-1=2. 1/{16}-1=-7/8` `\qquad sin^2a=1-cos^2a=1-1/{16}={15}/{16}` $\\$ `\qquad Q=sin^2 a+sin(π/ 6+a).sin(π/6-a)+cos(2a+5π)` `=sin^2 a+ 1/2. [cos(π/6+a-π/6+a)-cos(π/6+a+π/6-a)]+cos(2a+π+4π)` `=sin^2 a +1/ 2 cos2a – 1/ 2 cos \ π/3 +cos(2a+π)` `={15}/{16}+1/ 2 cos 2a-1/ 2 . 1/ 2 -cos2a` `={15}/{16}-1/ 4 -1/ 2 cos2a` `={11}/{16}-1/2 . {-7}/8=9/8` Vậy `Q=9/8` Bình luận
$Q=\sin^2a+\sin\Big(\dfrac{\pi}{6}+a\Big).\sin\Big(\dfrac{\pi}{6}-a\Big)+\cos(2a+5\pi)$ $=\sin^2a-\dfrac{1}{2}.\Big(cos\dfrac{\pi}{3}-\cos2a\Big)+\cos(\pi+2a)$ $=1-\cos^2a-\dfrac{1}{4}+\dfrac{1}{2}\cos2a-\cos2a$ $=1-\cos^2a-\dfrac{1}{4}+\dfrac{1}{2}(2\cos^2a-1)-2\cos^2a+1$ $=-2\cos^2a+\dfrac{5}{4}$ Thay $\cos a=\dfrac{1}{4}\to Q=\dfrac{9}{8}$ Bình luận
Đáp án:
`Q=9/8`
Giải thích các bước giải:
Ta có:
`\qquad cosa=1/4`
`=>cos2a=2cos^2a-1=2. 1/{16}-1=-7/8`
`\qquad sin^2a=1-cos^2a=1-1/{16}={15}/{16}`
$\\$
`\qquad Q=sin^2 a+sin(π/ 6+a).sin(π/6-a)+cos(2a+5π)`
`=sin^2 a+ 1/2. [cos(π/6+a-π/6+a)-cos(π/6+a+π/6-a)]+cos(2a+π+4π)`
`=sin^2 a +1/ 2 cos2a – 1/ 2 cos \ π/3 +cos(2a+π)`
`={15}/{16}+1/ 2 cos 2a-1/ 2 . 1/ 2 -cos2a`
`={15}/{16}-1/ 4 -1/ 2 cos2a`
`={11}/{16}-1/2 . {-7}/8=9/8`
Vậy `Q=9/8`
$Q=\sin^2a+\sin\Big(\dfrac{\pi}{6}+a\Big).\sin\Big(\dfrac{\pi}{6}-a\Big)+\cos(2a+5\pi)$
$=\sin^2a-\dfrac{1}{2}.\Big(cos\dfrac{\pi}{3}-\cos2a\Big)+\cos(\pi+2a)$
$=1-\cos^2a-\dfrac{1}{4}+\dfrac{1}{2}\cos2a-\cos2a$
$=1-\cos^2a-\dfrac{1}{4}+\dfrac{1}{2}(2\cos^2a-1)-2\cos^2a+1$
$=-2\cos^2a+\dfrac{5}{4}$
Thay $\cos a=\dfrac{1}{4}\to Q=\dfrac{9}{8}$