Could I have some help solving this pleaseeee The answer is pi/8 apparently
Scrambled New Member Joined Sep 16, 2023 Messages 17 Location Sydney Gender Male HSC 2025 Jun 7, 2024 #1 Could I have some help solving this pleaseeee The answer is pi/8 apparently
WeiWeiMan Well-Known Member Joined Aug 7, 2023 Messages 913 Location behind you Gender Male HSC 2026 Jun 8, 2024 #2 Scrambled said: Could I have some help solving this pleaseeee The answer is pi/8 apparently View attachment 43352 Click to expand... tan^-1(√2-1)+tan^-1[(1-(√2-1))/(1+√2-1)] =π/4 tan^-1(√2-1)+tan^-1[(2-√2)/√2]=π/4 tan^-1(√2-1)+tan^-1(√2-1)=π/4 2tan^-1(√2-1)=π/4 tan^-1(√2-1)=π/8
Scrambled said: Could I have some help solving this pleaseeee The answer is pi/8 apparently View attachment 43352 Click to expand... tan^-1(√2-1)+tan^-1[(1-(√2-1))/(1+√2-1)] =π/4 tan^-1(√2-1)+tan^-1[(2-√2)/√2]=π/4 tan^-1(√2-1)+tan^-1(√2-1)=π/4 2tan^-1(√2-1)=π/4 tan^-1(√2-1)=π/8
Scrambled New Member Joined Sep 16, 2023 Messages 17 Location Sydney Gender Male HSC 2025 Jun 8, 2024 #3 WeiWeiMan said: tan^-1(√2-1)+tan^-1[(1-(√2-1))/(1+√2-1)] =π/4 tan^-1(√2-1)+tan^-1[(2-√2)/√2]=π/4 tan^-1(√2-1)+tan^-1(√2-1)=π/4 2tan^-1(√2-1)=π/4 tan^-1(√2-1)=π/8 Click to expand... ohhhh so there was a typo it was meant to be arctan dang. Anyways thank you! Last edited: Jun 8, 2024
WeiWeiMan said: tan^-1(√2-1)+tan^-1[(1-(√2-1))/(1+√2-1)] =π/4 tan^-1(√2-1)+tan^-1[(2-√2)/√2]=π/4 tan^-1(√2-1)+tan^-1(√2-1)=π/4 2tan^-1(√2-1)=π/4 tan^-1(√2-1)=π/8 Click to expand... ohhhh so there was a typo it was meant to be arctan dang. Anyways thank you!