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Question Number 75131 by mathocean1 last updated on 07/Dec/19

please help me to show that tan^2 ((π/8))+2tan((π/8))−1=0

$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{2tan}\left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{1}=\mathrm{0} \\ $$

Answered by mr W last updated on 07/Dec/19

tan (π/4)=1 ⇒tan 2×(π/8)=1 ⇒((2 tan (π/8))/(1−tan^2 (π/8)))=1 ⇒2 tan (π/8)=1−tan^2 (π/8) ⇒tan^2 (π/8)+2 tan (π/8)−1=0

$$\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{2}×\frac{\pi}{\mathrm{8}}=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\pi}{\mathrm{8}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{8}}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}\:\mathrm{tan}\:\frac{\pi}{\mathrm{8}}=\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{8}}+\mathrm{2}\:\mathrm{tan}\:\frac{\pi}{\mathrm{8}}−\mathrm{1}=\mathrm{0} \\ $$

Commented by mathocean1 last updated on 07/Dec/19

thank you sir...can you help me to deduct tan(π/8)?

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}...\mathrm{can}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\: \\ $$$$\mathrm{deduct}\:\mathrm{tan}\frac{\pi}{\mathrm{8}}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by peter frank last updated on 07/Dec/19

tan 2θ=((2tan θ)/(1−tan^2 θ)) but θ=(π/8) tan 2(π/8)=((2tan (π/8))/(1−tan^2 (π/8))) tan (π/4)=((2tan (π/8))/(1−tan^2 (π/8)))

$$\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta} \\ $$$${but}\:\:\theta=\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{tan}\:\mathrm{2}\frac{\pi}{\mathrm{8}}=\frac{\mathrm{2tan}\:\frac{\pi}{\mathrm{8}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \\ $$$$\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\frac{\mathrm{2tan}\:\frac{\pi}{\mathrm{8}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \\ $$$$ \\ $$$$ \\ $$

Commented by MJS last updated on 07/Dec/19

right idea, just put θ=(α/2) tan α =((2tan (α/2))/(1−tan^2 (α/2))) now solve for tan (α/2) tan (α/2) =−((1±(√(1+tan^2 α)))/(tan α)) now α=(π/4) ⇒ tan α =1 ⇒ tan (π/8) =−1+(√2) knowing tan (π/8) >0 we cannot use the second solution −1−(√2)

$$\mathrm{right}\:\mathrm{idea},\:\mathrm{just}\:\mathrm{put}\:\theta=\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha\:=\frac{\mathrm{2tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}} \\ $$$$\mathrm{now}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:=−\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}}{\mathrm{tan}\:\alpha} \\ $$$$\mathrm{now}\:\alpha=\frac{\pi}{\mathrm{4}}\:\Rightarrow\:\mathrm{tan}\:\alpha\:=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{tan}\:\frac{\pi}{\mathrm{8}}\:=−\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\mathrm{knowing}\:\mathrm{tan}\:\frac{\pi}{\mathrm{8}}\:>\mathrm{0}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{use}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{solution}\:−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$

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