Find The Value Of 8
Example 27 - Chapter 3 Class xi Trigonometric Functions (Term two)
Last updated at February. xiii, 2020 by Teachoo
Transcript
Instance 27 find the value of tan π/8. tan π/8 Putting Ο = 180Β° = tan (180Β°)/8 = tan (45Β°)/ii We detect tan (45Β°)/2 using tan 2x formula tan 2x = (2 tanβ‘π₯)/(1 βπ‘ππ2π₯) Putting ten = (45Β°)/2 tan ("2 Γ " (45Β°)/two) = (two tanβ‘γ (45Β°)/twoγ)/(1 βπ‘ππ2 (45Β°)/ii) tan 45Β° = (two tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππtwo (45Β°)/2) tan 45Β° = (ii tanβ‘γ (45Β°)/iiγ)/(i βπ‘ππ2 (45Β°)/2) 1 = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππii (45Β°)/two) 1 β tan2 (45Β°)/2 = 2tan (45Β°)/two (As tan 45Β° = 1) Let tan (ππΒ°)/π = x So, our equation becomes 1 β x2 = 2x 0 = 2x + x2 β ane x2 + 2x β 1 = 0 The higher up equation is of the form ax2 + bx + c = 0 where a = one, b = 2, c = βi Solution are x = (β π Β± β(πii β4ππ) )/twoπ = (β 2 Β± β((βii)two β 4 Γ one Γ (β1)) )/(2 Γ 1) = (βtwo Β± β(4 + 4))/2 = (βii Β± β8)/2 = (β2 Β± β(2 Γ ii Γ ii))/2 = (βtwo Β± twoβ2)/2 = (two ( β1 Β±β2 ))/2 = βi Β± β2 Thus, ten = βone Β± βtwo tan (45Β°)/ii = β1 Β± βtwo But tan (45Β°)/2 = βi β βtwo is not possible as (45Β°)/2 = 22.fiveΒ° lies in first quadrant & tan is positive in showtime quadrant Therefore, tan (45Β°)/ii = β1 + β2 i.east. tan π /π = βπ β ane
Find The Value Of 8,
Source: https://www.teachoo.com/2231/586/Example-27---Find-value-of-tan-pi-8---Chapter-3-Class-11/category/Examples/
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