cos平方a分之1-tan平方a化簡
原式=1/cos²a-sin²/cos²a
=(1-sin²a)/cos²a
=cos²a/cos²a
=1
幫我化簡,cos平方A乘tan平方A+cos平方A
cos^2atan^2a+cos^2a
=sin^2a+cos^2a
=1
化簡cos平方x(1-tan平方x)
cos平方x(1-tan平方x)
=cos²x·(1-sin²x/cos²x)
=cos²x-sin²x
=cos2x
已知tanθ=1 3,則cos2θ+1 2sin2θ= ___.
∵tanθ=1
3,
∴cos2θ+1
2sin2θ=cos2θ+sinθcosθ
cos2θ+sin2θ=1+tanθ
1+tan2θ=1+1
3
1+1
9=6
5,
故答案為:6
5.
已知tanα=1,求2sin2α-3sinαcos-5cos2α的值 要步驟噢~拜託lo~
∵tana=1∴cosa=sina∴2sin2α-3sinαcosα-5cos2α=(2sin²a-3sinacosa-5cos²a)/(sin²a+cos²a)=(2sin²a-3sin²a-5sin²a)/(sin²a+sin²a)= -6sin²a/(2sin²a)…
若1/tanα=3,則cos²α+1/2sian2α=? 求詳解.
cos²α+1/2sian2α
=(cos²α+sinacosa)/(cos^2a+sin^2a)
=(1+cota)/(1+cot^2a)
=2/5
化簡:1-tanα分之-1+tanα分之1
(1/-1+tana)/1-tana(括弧內代表分子這一個分式)
=[1/-(1-tana)]/1-tana
=-1/(1-tana)²(同乘1-tana)
=-1/1-2tana+tana²(完全平方公式)
如果原題未錯,應該是這麼做.
化簡:1+tanα/1-tanα
1+tanα/1-tanα
=[tan(π/4)+tanα]/1-tanαtan(π/4)
=tan(α+π/4)
化簡tan+1/tan
tana+1/tana
=sina/cosa+coa/sina
=(sina²+cosa²)/sinacosa
=1/sinacosa
=2/sin2a
化簡tan((π/12)+1)/((π/12)-1)求指導,
題目有誤
可能是
[ tan(π/12)+1 ] / [ tan(π/12)- 1 ]
= [ tan(π/12)+ tan(π/4)] / [ tan(π/12)*tan(π/4)- 1 ]
= - tan(π/12 +π/4)
= - tan(π/3)
= -√3