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ta thấy cos12o = sin78o nên cos212o = sin278o
Sau khi biến đổi ta được: S= sin278o + cos278o + sin289o + cos289o=2
đáp án C
`sin^2x+cos^2x=1`
`<=>sin^2x+(1/2)^2=1`
`<=> sinx=\pm \sqrt3/2`
• `sinx=\sqrt3/2 => P=3. (\sqrt3/2)^2 +1=13/4`
• `sinx=-\sqrt3/2 => P = 3.(-\sqrt3/2) +1=13/4`
`=>` A.
\(P=3sin^2x+1=3\left(1-cos^2x\right)+1=3\left(1-\dfrac{1}{4}\right)+1=\dfrac{13}{4}\)
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)
b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)
c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)
d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
\(P=sin^2x+3cos^2x=1-cos^2x+3cos^2x=1+2cos^2x=1+2.\left(\dfrac{1}{4}\right)^2=\dfrac{9}{8}\)
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
f/
\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)
\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)
\(=4sinC.sinA.sinB\)
g/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)
\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(=1-cosC.cos\left(A-B\right)+cos^2C\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=1-2cosC.cosA.cosB\)
d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)
\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)
e/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
S = cos 2 13 ° + cos 2 32 ° + cos 2 58 ° + cos 2 77 ° = cos 2 13 ° + cos 2 32 ° + cos 2 ( 90 - 32 ) ° + cos 2 ( 90 - 13 ) ° = c o s 2 13 ° + cos 2 32 ° + sin 2 32 ° + sin 2 13 ° = 2
Chọn B.