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QUESTION:
If Sin θ + Cos θ = m, then prove that Sin⁶ θ Cos⁶ θ = [4 - 3(m² - 1)²] / 4, where m² ≤ 2.
GIVEN:
TO PROVE:
[tex]\sf Sin^6 \ \theta + Cos^6 \ \theta = \dfrac{3-4(m^2-1)^2}{4}[/tex]
EXPLANATION:
[tex]\sf \sf Sin^6 \ \theta + Cos^6 \ \theta = (Sin^2 \ \theta)^3 + (Cos^2 \ \theta)^3[/tex]
[tex]\boxed{\bold{\gray{A^3 + B^3 = (A + B)(A^2 - AB + B^2)}}}[/tex]
[tex]\sf \leadsto (Sin^2 \ \theta +Cos^2 \ \theta)(Sin^4 \ \theta -Sin^2 \ \theta\ Cos^2 \ \theta+Cos^4 \ \theta)[/tex]
[tex]\boxed{\bold{\gray{\large{Sin^2 \ \theta +Cos^2 \ \theta=1}}}}[/tex]
[tex]\sf \leadsto (1)(Sin^4 \ \theta -Sin^2 \ \theta \ Cos^2 \ \theta+Cos^4 \ \theta)[/tex]
[tex]\sf \leadsto Sin^4 \ \theta +Cos^4 \ \theta =(Sin^2 \ \theta +Cos^2 \ \theta )^2- 2\ Sin^2 \ \theta \ Cos^2 \ \theta[/tex]
[tex]\boxed{\bold{\gray{\large{Sin^2 \ \theta +Cos^2 \ \theta=1}}}}[/tex]
[tex]\sf \leadsto Sin^4 \ \theta +Cos^4 \ \theta =(1)^2- 2 \ Sin^2 \ \theta \ Cos^2 \ \theta[/tex]
[tex]\sf \leadsto Sin^4 \ \theta +Cos^4 \ \theta =1-2 \ Sin^2 \ \theta \ Cos^2 \ \theta[/tex]
[tex]\sf Subtract \ by \ sin^2 \ \theta \ cos^2 \ \theta[/tex]
[tex]\sf \leadsto Sin^4 \ \theta -Sin^2 \ \theta \ Cos^2 \ \theta+Cos^4 \ \theta = 1-3 \ Sin^2 \ \theta \ Cos^2 \ \theta[/tex]
[tex]\sf Sin \ \theta + Cos \ \theta = m[/tex]
Square on both sides.
[tex]\boxed{\bold{\large{\gray{(A + B)^2 = A^2 +2AB +B^2}}}}[/tex]
[tex]\sf Sin^2 \ \theta + Cos^2 \ \theta+2\ sin \ \theta\ cos \ \theta = m^2[/tex]
[tex]\boxed{\bold{\gray{\large{Sin^2 \ \theta +Cos^2 \ \theta=1}}}}[/tex]
[tex]\sf 1+2\ sin \ \theta\ cos \ \theta = m^2[/tex]
[tex]\sf 2\ sin \ \theta\ cos \ \theta = m^2-1[/tex]
[tex]\sf sin \ \theta\ cos \ \theta = \dfrac{m^2-1}{2}[/tex]
Square on both sides
[tex]\sf sin^2 \ \theta\ cos^2\ \theta = \dfrac{(m^2-1)^2}{4}[/tex]
Substitute the value of Sin² θ + Cos² θ
[tex]\sf 1-3 \dfrac{(m^2-1)^2}{4}[/tex]
[tex]\sf \dfrac{4}{4} -3 \dfrac{(m^2-1)^2}{4}[/tex]
[tex]\sf \dfrac{4-3(m^2-1)^2}{4}[/tex]
[tex]\sf Sin^6 \ \theta + Cos^6 \ \theta = \dfrac{3-4(m^2-1)^2}{4}[/tex]
HENCE PROVED.