giúp vs mk cần gấp lắm luôn sin^2 a + tan 45 độ + sin^2 a .cos^2 a + cos 60 độ 03/07/2021 Bởi Jasmine giúp vs mk cần gấp lắm luôn sin^2 a + tan 45 độ + sin^2 a .cos^2 a + cos 60 độ
Đáp án: $\begin{array}{l}{\sin ^2}a + \tan {45^0} + {\sin ^2}a.{\cos ^2}a + \cos {60^0}\\ = {\sin ^2}a + {\sin ^2}a.{\cos ^2}a + \tan {45^0} + \cos {60^0}\\ = {\sin ^2}a\left( {1 + {{\cos }^2}a} \right) + 1 + \dfrac{1}{2}\\ = \left( {1 – {{\cos }^2}a} \right)\left( {1 + {{\cos }^2}a} \right) + \dfrac{3}{2}\\ = 1 – {\cos ^4}a + \dfrac{3}{2}\\ = \dfrac{5}{2} – {\cos ^4}a\end{array}$ Bình luận
$\sin^2a+\tan45^o+\sin^2a.\cos^2a+\cos60^o$ $=\sin^2a+1+\sin^2a.\cos^2a+\dfrac{1}{2}$ $=\sin^2a(1+\cos^2a)+\dfrac{3}{2}$ Bình luận
Đáp án:
$\begin{array}{l}
{\sin ^2}a + \tan {45^0} + {\sin ^2}a.{\cos ^2}a + \cos {60^0}\\
= {\sin ^2}a + {\sin ^2}a.{\cos ^2}a + \tan {45^0} + \cos {60^0}\\
= {\sin ^2}a\left( {1 + {{\cos }^2}a} \right) + 1 + \dfrac{1}{2}\\
= \left( {1 – {{\cos }^2}a} \right)\left( {1 + {{\cos }^2}a} \right) + \dfrac{3}{2}\\
= 1 – {\cos ^4}a + \dfrac{3}{2}\\
= \dfrac{5}{2} – {\cos ^4}a
\end{array}$
$\sin^2a+\tan45^o+\sin^2a.\cos^2a+\cos60^o$
$=\sin^2a+1+\sin^2a.\cos^2a+\dfrac{1}{2}$
$=\sin^2a(1+\cos^2a)+\dfrac{3}{2}$