sin α – sin α .cos ²α ? tan ²α (2cos ²α +sin ²α -1)? 01/10/2021 Bởi Samantha sin α – sin α .cos ²α ? tan ²α (2cos ²α +sin ²α -1)?
Đáp án: Giải thích các bước giải: $\begin{array}{l} + ) = \sin a\left( {1 – {{\cos }^2}a} \right) = \sin a.{\sin ^2}a = {\sin ^3}a\\ + ) = {\tan ^2}a\left( {{{\cos }^2}a + {{\sin }^2}a – 1 + {{\cos }^2}a} \right)\\ = \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}\left( {1 – 1 + {{\cos }^2}a} \right)\\ = \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}.{\cos ^2}a\\ = {\sin ^2}a \end{array}$ Bình luận
$sin\alpha – sin\alpha.cos^2\alpha$ $= sin\alpha (1-cos^2\alpha)$ $= sin\alpha.sin^2\alpha$ $=sin^3\alpha$ $tan^2\alpha.(cos^2\alpha+cos^2\alpha+sin^2\alpha -1)$ $= tan^2\alpha.cos^2\alpha$ $= sin^2\alpha$ Bình luận
Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
+ ) = \sin a\left( {1 – {{\cos }^2}a} \right) = \sin a.{\sin ^2}a = {\sin ^3}a\\
+ ) = {\tan ^2}a\left( {{{\cos }^2}a + {{\sin }^2}a – 1 + {{\cos }^2}a} \right)\\
= \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}\left( {1 – 1 + {{\cos }^2}a} \right)\\
= \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}.{\cos ^2}a\\
= {\sin ^2}a
\end{array}$
$sin\alpha – sin\alpha.cos^2\alpha$
$= sin\alpha (1-cos^2\alpha)$
$= sin\alpha.sin^2\alpha$
$=sin^3\alpha$
$tan^2\alpha.(cos^2\alpha+cos^2\alpha+sin^2\alpha -1)$
$= tan^2\alpha.cos^2\alpha$
$= sin^2\alpha$