Question

If sinx + cosx = root 7 by 2 where x is member of [ 0, pi by 4] , then find the value of tan x by 2

Answer 1

given, sinx + cosx = √7/2 , where x $\in$ [0, π/4 ]

using formula, sin2A = 2tanA/(1 + tan²A)

using formula, sin2A = 2tanA/(1 + tan²A)cos2A = (1 - tan²A)/(1 + tan²A)

so, sinx = 2tan(x/2)/{1 + tan²(x/2)}

cosx = (1 - tan²(x/2))/{1 + tan²(x/2)}

so, 2tan(x/2)/{1 + tan²(x/2)} + {1 - tan²(x/2)}/{1 + tan²(x/2)} = √7/2

or, {1 - tan²(x/2) + 2tan(x/2)}/{1 + tan²(x/2)} = √7/2

or, 2 - 2tan²(x/2) + 4tan(x/2) = √7 + √7tan²(x/2)

or, (√7 + 2)tan²(x/2) - 4tan(x/2) + (√7 - 2) = 0

Let tan(x/2) = p

or, (√7 + 2)p² - 4p + (√7 - 2) = 0

or, p = {4 ± √{16 - 12}/2(√7 + 2)

= (2 ± 1 )/(√7 + 2)

= 1/(√7 + 2) , 3/(√7 + 2)

= (√7 - 2)/3 , (√7 - 2)

hence, tan(x/2) = (√7 - 2)/3 or, (√7 - 2)

but x lies between [0, π/4 ]

so, x/2 lies between [0, π/8] , tan(π/8) ≈ 0.41

so, tan(x/2) ≠ (√7 - 2)

hence, value of tan(x/2) = (√7 - 2)/3

Answer 2

Answer:

given, sinx + cosx = √7/2 , where x \in∈ [0, π/4 ]

using formula, sin2A = 2tanA/(1 + tan²A)

using formula, sin2A = 2tanA/(1 + tan²A)cos2A = (1 - tan²A)/(1 + tan²A)

so, sinx = 2tan(x/2)/{1 + tan²(x/2)}

cosx = (1 - tan²(x/2))/{1 + tan²(x/2)}

so, 2tan(x/2)/{1 + tan²(x/2)} + {1 - tan²(x/2)}/{1 + tan²(x/2)} = √7/2

or, {1 - tan²(x/2) + 2tan(x/2)}/{1 + tan²(x/2)} = √7/2

or, 2 - 2tan²(x/2) + 4tan(x/2) = √7 + √7tan²(x/2)

or, (√7 + 2)tan²(x/2) - 4tan(x/2) + (√7 - 2) = 0

Step-by-step explanation:

Let tan(x/2) = p

or, (√7 + 2)p² - 4p + (√7 - 2) = 0

or, p = {4 ± √{16 - 12}/2(√7 + 2)

= (2 ± 1 )/(√7 + 2)

= 1/(√7 + 2) , 3/(√7 + 2)

= (√7 - 2)/3 , (√7 - 2)

hence, tan(x/2) = (√7 - 2)/3 or, (√7 - 2)

but x lies between [0, π/4 ]

so, x/2 lies between [0, π/8] , tan(π/8) ≈ 0.41

so, tan(x/2) ≠ (√7 - 2)

hence, value of tan(x/2) = (√7 - 2)/3