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Question Number 128808 by liberty last updated on 10/Jan/21

If x cos q − sin q = 1 then x^2 +(1+x^2 )sin q =?

$$\:\mathrm{If}\:{x}\:\mathrm{cos}\:{q}\:−\:\mathrm{sin}\:{q}\:=\:\mathrm{1}\:\mathrm{then}\:{x}^{\mathrm{2}} +\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mathrm{sin}\:{q}\:=?\: \\ $$

Commented by benjo_mathlover last updated on 10/Jan/21

x = ((cos^2 (q/2)+sin^2 (q/2)+2sin (q/2)cos (q/2))/(cos^2 (q/2)−sin^2 (q/2))) = ((cos (q/2)+sin (q/2))/(cos (q/2)−sin (q/2))) x^2 =((1+sin q)/(1−sin q)) then 1+x^2 = ((1+sin q)/(1−sin q)) + 1= (2/(1−sin q)) Now x^2 +(1+x^2 )sin q = ((1+sin q)/(1−sin q)) + ((2sin q)/(1−sin q)) = ((1+3sin q)/(1−sin q))

$$\:{x}\:=\:\frac{\mathrm{cos}\:^{\mathrm{2}} \frac{{q}}{\mathrm{2}}+\mathrm{sin}\:^{\mathrm{2}} \frac{{q}}{\mathrm{2}}+\mathrm{2sin}\:\frac{{q}}{\mathrm{2}}\mathrm{cos}\:\frac{{q}}{\mathrm{2}}}{\mathrm{cos}\:^{\mathrm{2}} \frac{{q}}{\mathrm{2}}−\mathrm{sin}\:^{\mathrm{2}} \frac{{q}}{\mathrm{2}}}\:=\:\frac{\mathrm{cos}\:\frac{{q}}{\mathrm{2}}+\mathrm{sin}\:\frac{{q}}{\mathrm{2}}}{\mathrm{cos}\:\frac{{q}}{\mathrm{2}}−\mathrm{sin}\:\frac{{q}}{\mathrm{2}}} \\ $$$$\:{x}^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{sin}\:{q}}{\mathrm{1}−\mathrm{sin}\:{q}}\:\:{then}\:\mathrm{1}+{x}^{\mathrm{2}} \:=\:\frac{\mathrm{1}+\mathrm{sin}\:{q}}{\mathrm{1}−\mathrm{sin}\:{q}}\:+\:\mathrm{1}=\:\frac{\mathrm{2}}{\mathrm{1}−\mathrm{sin}\:{q}} \\ $$$${N}\mathrm{ow}\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{sin}\:\mathrm{q}\:=\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{q}}{\mathrm{1}−\mathrm{sin}\:\mathrm{q}}\:+\:\frac{\mathrm{2sin}\:\mathrm{q}}{\mathrm{1}−\mathrm{sin}\:\mathrm{q}}\:=\:\frac{\mathrm{1}+\mathrm{3sin}\:\mathrm{q}}{\mathrm{1}−\mathrm{sin}\:\mathrm{q}} \\ $$

Answered by bramlexs22 last updated on 10/Jan/21

⇔ x = ((1+sin q)/(cos q)) ; x^2 = ((1+2sin q+sin^2 q)/(cos^2 q)) ⇔ 1+x^2 = ((cos^2 q+1+2sin q+sin^2 q)/(cos^2 q)) ⇔ ((2+2sin q)/(cos^2 q)) then x^2 +(1+x^2 )sin q = ((1+2sin q+sin^2 q)/(cos^2 q)) + (((2+2sin q)/(cos^2 q))).sin q= ((1+4sin q+3sin^2 q)/(cos^2 q)) = ((4sin^2 q+4sin q+cos^2 q)/(cos^2 q)) = 1+((4sin q(sin q+1))/(1−sin^2 q)) = 1+ ((4sin q)/(1−sin q)) = 1+ (4/((1/(sin q)) −1))

$$\:\Leftrightarrow\:\mathrm{x}\:=\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{q}}{\mathrm{cos}\:\mathrm{q}}\:;\:\mathrm{x}^{\mathrm{2}} =\:\frac{\mathrm{1}+\mathrm{2sin}\:\mathrm{q}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{q}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}} \\ $$$$\Leftrightarrow\:\mathrm{1}+\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}+\mathrm{1}+\mathrm{2sin}\:\mathrm{q}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{q}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}} \\ $$$$\:\Leftrightarrow\:\frac{\mathrm{2}+\mathrm{2sin}\:\mathrm{q}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}}\:\mathrm{then}\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{sin}\:\mathrm{q}\:= \\ $$$$\frac{\mathrm{1}+\mathrm{2sin}\:\mathrm{q}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{q}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}}\:+\:\left(\frac{\mathrm{2}+\mathrm{2sin}\:\mathrm{q}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}}\right).\mathrm{sin}\:\mathrm{q}= \\ $$$$\:\frac{\mathrm{1}+\mathrm{4sin}\:\mathrm{q}+\mathrm{3sin}\:^{\mathrm{2}} \mathrm{q}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}}\:=\:\frac{\mathrm{4sin}\:^{\mathrm{2}} \mathrm{q}+\mathrm{4sin}\:\mathrm{q}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{q}} \\ $$$$\:=\:\mathrm{1}+\frac{\mathrm{4sin}\:\mathrm{q}\left(\mathrm{sin}\:\mathrm{q}+\mathrm{1}\right)}{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{q}}\: \\ $$$$\:=\:\mathrm{1}+\:\frac{\mathrm{4sin}\:\mathrm{q}}{\mathrm{1}−\mathrm{sin}\:\mathrm{q}}\:=\:\mathrm{1}+\:\frac{\mathrm{4}}{\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{q}}\:−\mathrm{1}} \\ $$$$ \\ $$

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