函数和几何题,
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函数和几何题,
![](http://img.wesiedu.com/upload/a/e1/ae143425d586cf1052df0e34a42fb255.jpg)
![](http://img.wesiedu.com/upload/b/05/b0578cbba9deb46e1fd9ef53856df899.jpg)
![](http://img.wesiedu.com/upload/a/e1/ae143425d586cf1052df0e34a42fb255.jpg)
![](http://img.wesiedu.com/upload/b/05/b0578cbba9deb46e1fd9ef53856df899.jpg)
![函数和几何题,](/uploads/image/z/2932687-55-7.jpg?t=%E5%87%BD%E6%95%B0%E5%92%8C%E5%87%A0%E4%BD%95%E9%A2%98%2C)
1:解析1),由题意知sina=3/5,tana=sina/cosa=-3/4,
2),原式=4tana+3=-3+3=0
2,1)由图像可知,A=2,T/2=11π/12-5π/12=π/2,得T=π,∴ω=2π/T=2π/π=2,
则y=2sin(2x+φ)
又f(11π/12)=2sin(2*11π/12+φ)=0,解得φ=π/6,
∴y=2sin(2x+π/6),2),单调递增区间为(kπ-π/3,kπ+π/6),单调递减区间为(kπ+π6,kπ+2π/3),
3),当x∈[0,π/2]时,2x+π/6∈[π/6,7π/6],y∈[-1,2],即f(x)取值范围为【-1,2】
3,证明:1)提示,易证AD⊥BC,又直可得B1B⊥AD,得AD⊥面BCC1B1,最后得结论
2)连接Ac1和A1C,交与O,由中位线定理易证OD∥A1B,结论得证.那里不懂喊我或留言.
2),原式=4tana+3=-3+3=0
2,1)由图像可知,A=2,T/2=11π/12-5π/12=π/2,得T=π,∴ω=2π/T=2π/π=2,
则y=2sin(2x+φ)
又f(11π/12)=2sin(2*11π/12+φ)=0,解得φ=π/6,
∴y=2sin(2x+π/6),2),单调递增区间为(kπ-π/3,kπ+π/6),单调递减区间为(kπ+π6,kπ+2π/3),
3),当x∈[0,π/2]时,2x+π/6∈[π/6,7π/6],y∈[-1,2],即f(x)取值范围为【-1,2】
3,证明:1)提示,易证AD⊥BC,又直可得B1B⊥AD,得AD⊥面BCC1B1,最后得结论
2)连接Ac1和A1C,交与O,由中位线定理易证OD∥A1B,结论得证.那里不懂喊我或留言.