3sinα+4cosα=5,则tanα=

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3sinα+4cosα=5,则tanα=3sinα+4cosα=5,则tanα=3sinα+4cosα=5,则tanα=两边同时除以5得(3/5)sina+(4/5)cosa=1cos53°*sina

3sinα+4cosα=5,则tanα=
3sinα+4cosα=5,则tanα=

3sinα+4cosα=5,则tanα=
两边同时除以 5 得
(3/5)sina+(4/5)cosa=1
cos53°*sina+sin53°*cosa=1
sin(53°+a)=1
a=37°+2kπ
tana=3/4

25(cos^2a+sin^2a)=9sin^2a+16cos^2a+24sinacosa
16sin^2a+9cos^2a-24sinacosa=0
(4sina-3cosa)^2=0
4sina=3cosa
tana=3/4