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Question Number 139521 by mnjuly1970 last updated on 28/Apr/21
              #   nice ... calculus#          if    (pq)^(sin^2 (x)) +(pq)^(cos^2 (x)) =p+q            then    tan(x)=?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:#\:\:\:{nice}\:…\:{calculus}# \\ $$$$\:\:\:\:\:\:\:\:{if}\:\:\:\:\left({pq}\right)^{{sin}^{\mathrm{2}} \left({x}\right)} +\left({pq}\right)^{{cos}^{\mathrm{2}} \left({x}\right)} ={p}+{q} \\ $$$$\:\:\:\:\:\:\:\:\:\:{then}\:\:\:\:{tan}\left({x}\right)=? \\ $$
Answered by qaz last updated on 28/Apr/21
≪1≫:p=q=0   tan x=???  ≪2≫:p=q=1   tan x=???  ≪3≫:p,q∈(0,1)∪(1,∞),pq≠1  (pq)^(sin^2 x) +(pq)^(cos^2 x)   =(pq)^(sin^2 x) +(pq)^(1−sin^2 x) =p+q  ⇒   sin^2 x=log_((pq)) p     cos^2 x=log_((pq)) q  or   sin^2 x=log_((pq)) q      cos^2 x=log_((pq)) p  tan x=(√(log_p q))     or  (√(log_q p))  ≪4≫:p,q∈C   tan x=???  .....
$$\ll\mathrm{1}\gg:{p}={q}=\mathrm{0}\:\:\:\mathrm{tan}\:{x}=??? \\ $$$$\ll\mathrm{2}\gg:{p}={q}=\mathrm{1}\:\:\:\mathrm{tan}\:{x}=??? \\ $$$$\ll\mathrm{3}\gg:{p},{q}\in\left(\mathrm{0},\mathrm{1}\right)\cup\left(\mathrm{1},\infty\right),{pq}\neq\mathrm{1} \\ $$$$\left({pq}\right)^{\mathrm{sin}\:^{\mathrm{2}} {x}} +\left({pq}\right)^{\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$$=\left({pq}\right)^{\mathrm{sin}\:^{\mathrm{2}} {x}} +\left({pq}\right)^{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}} ={p}+{q} \\ $$$$\Rightarrow\:\:\:\mathrm{sin}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {p}\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {q} \\ $$$${or}\:\:\:\mathrm{sin}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {q}\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {p} \\ $$$$\mathrm{tan}\:{x}=\sqrt{{log}_{{p}} {q}}\:\:\:\:\:{or}\:\:\sqrt{{log}_{{q}} {p}} \\ $$$$\ll\mathrm{4}\gg:{p},{q}\in{C}\:\:\:\mathrm{tan}\:{x}=??? \\ $$$$….. \\ $$

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