Question

If sinA - cosA=1, find value of sin^4A+cos^4A

Answer 1

sinA-cosA=1
Squaring both sides,
(sinA-cosA)²=1
or, sin²A-2sinAcosA+cos²A=1
or, sin²A+cos²A-2sinAcosA=1
or, 1-2sinAcosA=1
or, -2sinAcosA=1-1
or, sinAcosA=0 --------------------(1)
sin⁴A+cos⁴A
=(sin²A)²+(cos²A)²
=(sin²A+cos²A)²-2sin²Acos²A
=1²-2(sinAcosA)²
=1-2(0)² [Using (1)]
=1 Ans.

Answer 2

Answer:

$sin^4A+cos^4A =1$

Step-by-step explanation:

We are given that $sinA-cosA=1$

Squaring both sides,

$(sinA-cosA)^2=1$

Identity : $(a-b)^2= a^2+b^2-2ab$

$sin^2A-2sinAcosA+cos^2A=1$

Since we know that $Sin^2A+Cos^2A=1$

$1-2sinAcosA=1$

$-2sinAcosA=0$

$sinAcosA=0$

We are supposed to find $sin^4A+cos^4A$

$(sin^2A)^2+(cos^2A)^2$

Using identity : $a^2+b^2+2ab = (a+b)^2$

=$(sin^2A+cos^2A)^2-2sin^2Acos^2A$

=$1^2-2(sinAcosA)^2$

=$1-2(0)^2$

=1

Hence $sin^4A+cos^4A =1$