\(\left(sinx+cosx\right)^2=\frac{25}{16}\Rightarrow sin^2x+cos^2x+2sinxcosx=\frac{25}{16}\)

\(\Rightarrow2sinxcosx=\frac{25}{16}-1=\frac{9}{16}\Rightarrow A=\frac{9}{32}\)

\(B^2=\left(sinx-cosx\right)^2=1-2sinx.cosx=1-\frac{9}{16}=\frac{7}{16}\Rightarrow B=\pm\frac{\sqrt{7}}{4}\)

\(C=\left(sinx+cosx\right)\left(sinx-cosx\right)=\frac{5}{4}.\left(\pm\frac{\sqrt{7}}{4}\right)=\pm\frac{5\sqrt{7}}{16}\)

c)

\(\cos\left(x\right)^4+\sin\left(x\right)^2\cos\left(x\right)^2+\sin\left(x\right)^2\\ =\left(\cos\left(x\right)^2+\sin\left(x\right)^2\right)\cos\left(x\right)^2+\sin\left(x\right)^2\\ =\cos\left(x\right)^2+\sin\left(x\right)^2\\ =1\)

\(\cos\left(x\right)^4-\sin\left(x\right)^4+2\sin\left(x\right)^2\\ =\left(\cos\left(x\right)^2-\sin\left(x\right)^2\right)\left(\cos\left(x\right)^2+\sin\left(x\right)^2\right)+2\sin\left(x\right)^2\\ =\cos\left(2x\right)\cdot1+2\sin\left(x\right)^2\\ =\cos\left(x\right)^2-\sin\left(x\right)^2+2\sin\left(x\right)^2\\ =\cos\left(x\right)^2+\sin\left(x\right)^2\\ =1\)

b)

\(\sin\left(x\right)^4+\sin\left(x\right)^2\cos\left(x\right)^2+\cos\left(x\right)^2\\ =\sin\left(x\right)^2\left(\sin\left(x\right)^2+\cos\left(x\right)^2\right)+\cos\left(x\right)^2\\ =\sin\left(x\right)^2+\cos\left(x\right)^2\\ =1\)

\(\sin^6x+\cos^6x\\ =\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\\ =\left(\sin^2x+\cos^2x\right)^3-3\sin^2x\cos^2x\left(\sin^2x+\cos^2x\right)\\ =1-3\sin^2x\cos^2x\left(đpcm\right)\)

\(sin^6x+cos^6x\)

=\(\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)

=\(sin^4x-sin^2x.cos^2x+cos^4x\)

=\(\left(1-2sin^2x.cos^2x\right)-sin^2x.cos^2x\)

=\(1-3sin^2x.cos^2x\)(đpcm)

➞\(sin^6x+cos^6x\)=\(1-3sin^2x.cos^2x\)

1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)

⇔  \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)

2, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx=1-\sqrt{3}\)

⇔ \(\dfrac{\left(\sqrt{3}-1\right)}{2\sqrt{2}}sinx+\dfrac{\left(\sqrt{3}+1\right)}{2\sqrt{2}}cosx=\dfrac{1-\sqrt{3}}{2\sqrt{2}}\)

⇔ sinx . si

Điều kiện góc cần tính là góc nhọn

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

\(\Rightarrow cos^2x=\frac{1}{1+tan^2x}\Rightarrow cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{4}{5}\)

\(\Rightarrow sinx=\sqrt{1-cos^2x}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\)

\(\Rightarrow A=\frac{7}{5}\)