Grade 11Trigonometry
Find the max and min value of cos*(45+x)+(sinx-cosx)* (*means square)
2 Answers
Vikas TU
cos*(45+x)+(sinx-cosx)*
open the square ones,
=> cos*(45+x)+((sinx)* + (cosx)* -2sinxcosx)
=> cos*(45+x) + 1 – sin2x
now, maximum value for cos^2x is 1. and minimum zero.
and sin2x would have -1 when cos^2x is 1 and 1 when cos^2x is 0.
Because sinx and cos cannot have maximum and minimum value simulataneously .
Ajay
The maximum value of the function is 3 and minimum vaues is 0 as proved below.
using formula for cos2x and expanding the bracket the expression becomes
1/2[cos(pie/2 + 2x) + 1] + 1 – 2sinxcosx
= 1/2[-sin2x + 1] + 1 -sin2x
= 3/2 -3/2sin2x
The maximum value is when sin2x = -1
max value = 3/2+ 3/2 = 3.
The mimimum value is when sin2x = 1
min value =3/2-3/2 = 0
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