Trigonometry - Find The Range of sin^4x + cos^4x

Trigonometry - Find The Range of sin^4x + cos^4x

[armacham]

Another way:
make a new variable u = sin^2
1 = sin^2 + cos^2
1 = u + cos^2
1 - u = cos^2

now you can substitute u into the original equation to get:

y = u^2 + (1-u)^2
y = 2u^2 - 2u + 1

we know the range of u is just [0, 1] because sine is between -1 and 1 so sin^2 can only be 0 to 1

so we know we have to test the edges (u=0 and u=1) but we also need to test whatever maxima/minima there are between 0 and 1

to find that, just take the derivative with respect to u

y' = 4u - 2

and set the derivative equal to zero

4u - 2 = 0
4u = 2
u = 0.5

so now we need to find y-values when u = 0, 0.5, and 1. and whatever those three values are, the biggest one is the top of the range and the smallest one is the bottom of the range.

y = 2u^2 - 2u + 1
if u = 0: y = 0 - 0 + 1 = 1
if u = 0.5: y = 2(1/4) - 2(1/2) + 1 = 0.5
if u = 1: y = 2 - 2 + 1 = 1

so we have two distinct values, 0.5 and 1
those must be the boundaries, so the range is [0.5, 1]

[vcvartak]

Upper bound 1 is obvious, but getting lower bound you did it very systematic way

[IIT_Or_NiT]

2:20 sir sin 2x ka square 0 kaise aa gya

[Kaka_Shi]

in line 2, what did you do?
there is no power in 'x' but u still did it, '(sin^2 x)^2