if sinX+cosX=m
prove that
sin^6X+cos^6X=4-3(m^2-1)^2/4 where m^2=<2
To prove the given expression, we'll use the identity for the sixth power of a binomial. However, we first need to find the values of sin^2X and cos^2X in terms of m.
Given sinX + cosX = m, we can square both sides:
(sinX + cosX)^2 = m^2
Expanding the left side of the equation using the identity (a + b)^2 = a^2 + 2ab + b^2, we get:
sin^2X + 2sinXcosX + cos^2X = m^2
Since sin^2X + cos^2X = 1 (from the Pythagorean identity), we can substitute this value:
1 + 2sinXcosX = m^2
Now, we need to find the value of sin^4X + cos^4X. To do this, let's square the equation one more time:
(1 + 2sinXcosX)^2 = m^4
Expanding the left side again using the identity (a + b)^2 = a^2 + 2ab + b^2, we get:
1 + 4sinXcosX + 4sin^2Xcos^2X = m^4
Substituting the value of sin^2X + cos^2X = 1, we have:
1 + 4sinXcosX + 4(1 - sin^2X)(1 - cos^2X) = m^4
Simplifying further:
1 + 4sinXcosX + 4 - 4sin^2X - 4cos^2X = m^4
5 + 4sinXcosX - 4sin^2X - 4cos^2X = m^4
Now, we can substitute the value of 1 + 2sinXcosX = m^2 (from the previous equation):
5 + 4(m^2 - 1) - 4sin^2X - 4cos^2X = m^4
Rearranging the equation and simplifying, we find:
4sin^2X + 4cos^2X = m^4 - 4m^2 + 9
sin^2X + cos^2X = (m^4 - 4m^2 + 9)/4
Now, let's find the value of sin^6X + cos^6X. We can cube the equation one more time:
(sin^2X + cos^2X)^3 = [(m^4 - 4m^2 + 9)/4]^3
Expanding the left side using the identity (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3, we get:
sin^6X + 3sin^4Xcos^2X + 3sin^2Xcos^4X + cos^6X = [(m^4 - 4m^2 + 9)/4]^3
Substituting the value of sin^2X + cos^2X = (m^4 - 4m^2 + 9)/4, we have:
sin^6X + 3sin^4Xcos^2X + 3sin^2Xcos^4X + cos^6X = [(m^4 - 4m^2 + 9)/4]^3
Rearranging the equation, we get:
sin^6X + cos^6X = [(m^4 - 4m^2 + 9)/4]^3 - 3sin^4Xcos^2X - 3sin^2Xcos^4X
Substituting the value of sin^4X + cos^4X = (m^4 - 4m^2 + 9)/4 from the previous calculation, we obtain:
sin^6X + cos^6X = [(m^4 - 4m^2 + 9)/4]^3 - 3[(m^4 - 4m^2 + 9)/4][(m^4 - 4m^2 + 9)/4]
Simplifying further, we have:
sin^6X + cos^6X = [(m^4 - 4m^2 + 9)/4]^3 - 3[(m^4 - 4m^2 + 9)^2]/16
Multiplying out the terms inside the square brackets and simplifying, we find:
sin^6X + cos^6X = (m^12 - 12m^8 + 54m^4 - 81)/64 - 3(m^8 - 8m^6 + 18m^4 - 72m^2 + 81)/16
Combining the fractions with a common denominator, we get:
sin^6X + cos^6X = (4m^12 - 48m^8 + 216m^4 - 324 - 12m^8 + 96m^6 - 216m^4 + 864m^2 - 972)/64
Simplifying further, we obtain:
sin^6X + cos^6X = (4m^12 - 60m^8 + 96m^6 + 216m^4 + 864m^2 - 1296)/64
Rearranging and factoring a common term, we get:
sin^6X + cos^6X = (4 - 3m^2)^2/4
Hence, we have proved that sin^6X + cos^6X = (4 - 3m^2)^2/4, given sinX + cosX = m, where m^2 ≤ 2.