Question

solve who is the brainliest here​

Answer 1

Answer:

Put cos²x = 1-sin²x

sin²x+(1-sin²x)²

= sin²x + 1 -2sin²x +sin⁴x

= 1 - sin²x + sin⁴x

= 1 - sin²x(1-sin²x)

= 1 - sin²x.cos²x

=1 - (sinx.cosx)²

= 1 - ¼(2sinx.cosx)²

2sinx.cosx = sin(2x)

= 1 -¼(sin²(2x))

Now to find range.

We know that range of sin(x) is from -1 to 1

-1 ≤ sin(x) ≤ 1

-1 ≤ sin(2x) ≤ 1

0 ≤ sin²(2x) ≤ 1

0 ≤ ¼sin²(2x) ≤ ¼

-¼ ≤ -¼sin²(2x) ≤ 0

1-¼ ≤ 1-¼sin²(2x) ≤ 1+0

¾ ≤ 1 - ¼sin²(2x) ≤ 1

¾ ≤ 1 A ≤ 1 (Proved)

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