Question Number 102587 by 1549442205 last updated on 10/Jul/20

Prove that  cot7.5°=(√2)+(√3)+(√4)+(√6)

Commented byPRITHWISH SEN 2 last updated on 10/Jul/20

a different approach  cot θ=((cos θ)/(sin θ)) = ((2cos 7θcos θ)/(2cos 7θsin θ))  = ((cos 8θ+cos 6θ)/(sin 8θ−sin 6θ))  put θ=7∙5  cot 7∙5= ((cos 60+cos 45)/(sin 60−sin 45))=(((1+(√2)))/(((√3)−(√2))))=(1+(√2))((√3)+(√2))  =(√2)+ (√3) +(√4) +(√6)

Answered by Dwaipayan Shikari last updated on 10/Jul/20

Method 1 )  cot7.5°=((cos7.5°)/(sin7.5))=((2cos^2 7.5°)/(sin15°))=((1+cos15°)/(sin15°))=((2cos15°+1+cos30°)/(sin30°))          cos15=(((√3)+1)/(2(√2)))                                                          =2((((√3)+1)/(√2))+1+(√3).(1/2))                                                                                                  =2((((√6)+(√2)+(√4)+(√3))/2))  so    cot7.5°=(√2)+(√3)+(√4)+(√6) (  proved)      Method 2  cos(π/(24))=(√((1/2)(1+cos(π/(12)))))=(√((1/2)(1+(((√3)+1)/(2(√2))))))=(1/2)(√((2(√2)+(√3)+1)/(√2)))                                                                                          (1/(2(√2)))(√(4+(√6)+(√2)))  sin(π/(24))=(1/(2(√2))).(√(4−(√6)−(√2)))  cot(π/(24))=(√((4+(√6)+(√2))/(4−(√6)−(√2))))=((4+(√6)+(√2))/(16−8−2(√(12)))) =(1/2).((4+(√6)+(√2))/(4−(√(12))))=(1/2).(((4+(√6)+(√2))(4+(√(12))))/4)    =(1/8)(16+4(√6)+4(√2)+4(√(12))+6(√2)+2(√(12)))  =2+(√(3/2))+(1/(√2))+(√3)+(3/(2(√2)))+((√3)/(2(√2)))  =(√4)+((2(√3)+2+2(√6)+3+(√3))/(2(√2)))=(√4)+(√2)+(√3)+(√6)

Commented bybemath last updated on 10/Jul/20

how ((cos 7.5^o )/(sin 7.5^o )) = ((2cos ^2 7.5^o )/(sin 7.5^o )) ?

Commented byDwaipayan Shikari last updated on 10/Jul/20

multiplying by 2cos7.5°

Commented byDwaipayan Shikari last updated on 10/Jul/20

It is sin 15°

Commented bybemath last updated on 10/Jul/20

oo sin 15^o . thank

Answered by bobhans last updated on 10/Jul/20

tan 15^o  = tan (2×7.5^o ) = ((2tan 7.5^o )/(1−tan ^2 (7.5^o )))  (1) tan 15^o =tan (45^o −30^o )=((1−(1/(√3)))/(1+(1/(√3)))) =  (((√3)−1)/((√3)+1)) = ((4−2(√3))/2) = 2−(√3)   (2) set tan 7.5^o  = x  (2−(√3))(1−x^2 )= 2x  (2−(√3))x^2 +2x−(2−(√3) )=0  x = ((−2+ (√(4+4(2−(√3))^2 )))/(2(2−(√3)))) =  ((−1+(√(1+7−4(√3))))/(2−(√3))) = ((−1+(√(8−2(√(12)))))/(2−(√3))) =  ((−1+(√6)−(√2))/(2−(√3))) = ((((√6)−(√2)−1)(2+(√3)))/1) =  2(√6)+3(√2)−2(√2)−(√6)−2−(√3)=  (√6)+(√2)−2−(√3) = tan 7.5^o