题目内容
(1)证明:sin4θ+sin2θcos2θ+cos2θ=1
(2)计算:sin
π+cos
π+tan(-
π).
(2)计算:sin
| 25 |
| 6 |
| 25 |
| 3 |
| 25 |
| 4 |
(1)证明:左边=sin4θ+sin2θcos2θ+cos2θ=sin2θ(sin2θ+cos2θ)+cos2θ=sin2θ+cos2θ=1=右边,
则原式成立;
(2)原式=sin(4π+
)+cos(8π+
)-tan(6π+
)=sin
+cos
-tan
=
+
-1=1-1=0.
则原式成立;
(2)原式=sin(4π+
| π |
| 6 |
| π |
| 3 |
| π |
| 4 |
| π |
| 6 |
| π |
| 3 |
| π |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
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