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10 tháng 5 2017

\(sin\left(\alpha+\dfrac{\pi}{6}\right)-cos\left(\alpha-\dfrac{2\pi}{3}\right)\)
\(=cos\left(\dfrac{\pi}{3}-\alpha\right)-cos\left(\dfrac{2\pi}{3}-\alpha\right)\)
\(=-sin\left(\dfrac{\pi}{2}-\alpha\right)sin\left(-\dfrac{\pi}{6}\right)\)
\(=cos\alpha.sin\dfrac{\pi}{6}\)\(=\dfrac{1}{3}.\dfrac{1}{2}=\dfrac{1}{6}\).

NV
18 tháng 4 2021

\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)

\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)

NV
13 tháng 4 2021

1.

\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)

\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)

\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)

\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)

\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)

\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)

\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)

\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)

NV
13 tháng 4 2021

2.

\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)

\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x\)

\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)

\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)

\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)

\(=tan^2a.cot^2b-2\)

25 tháng 7 2018

bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)

\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)

\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\) \(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)

26 tháng 4 2017

Giải bài 4 trang 155 SGK Đại Số 10 | Giải toán lớp 10

Giải bài 4 trang 155 SGK Đại Số 10 | Giải toán lớp 10

26 tháng 4 2018

\(a\) thuộc góc phần tư thứ III -> sin\(a\) < 0

+) sin\(a\)=-\(\sqrt{1-cos^2a}\)=-\(\sqrt{1-\left(\dfrac{-12}{13}\right)^2}\)=\(\dfrac{-5}{13}\)

\(cos2a=cos^2a-sin^2a\)=\(\left(\dfrac{-12}{13}\right)^2-\left(\dfrac{-5}{13}\right)^2=\dfrac{119}{169}\)

23 tháng 4 2022

nhường cho I don't know :))

23 tháng 4 2022

Nó bt đếch gì mà nhường :) ?

\(B=\sqrt{2}\left(sina-cosa\right)-\sqrt{2}\left(cosa+sina\right)\)

\(=\sqrt{2}\cdot\left(-2cosa\right)=-2\sqrt{2}\cdot\dfrac{1}{3}=-\dfrac{2\sqrt{2}}{3}\)

5 tháng 7 2021

\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)

\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)

\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)

\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)

\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)

\(=cosx-cosx+sin^2x+cos^2x+sinx\)

\(=1+sinx\)

\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)

\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)

\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)

\(=1+cosx\)

5 tháng 7 2021

bị bỏ gp chị nhắn tin vs mấy ad ấy, nhanh ko ấy mà chị =))