Question

please solve this question for me​

Answer 1

QUESTION:

If Sin θ + Cos θ = m, then prove that Sin⁶ θ Cos⁶ θ = [4 - 3(m² - 1)²] / 4, where m² ≤ 2.

GIVEN:

• Sin θ + Cos θ = m

• m² ≤ 2

TO PROVE:

$\sf Sin^6 \ \theta + Cos^6 \ \theta = \dfrac{3-4(m^2-1)^2}{4}$

EXPLANATION:

$\sf \sf Sin^6 \ \theta + Cos^6 \ \theta = (Sin^2 \ \theta)^3 + (Cos^2 \ \theta)^3$

$\boxed{\bold{\gray{A^3 + B^3 = (A + B)(A^2 - AB + B^2)}}}$

$\sf \leadsto (Sin^2 \ \theta +Cos^2 \ \theta)(Sin^4 \ \theta -Sin^2 \ \theta\ Cos^2 \ \theta+Cos^4 \ \theta)$

$\boxed{\bold{\gray{\large{Sin^2 \ \theta +Cos^2 \ \theta=1}}}}$

$\sf \leadsto (1)(Sin^4 \ \theta -Sin^2 \ \theta \ Cos^2 \ \theta+Cos^4 \ \theta)$

$\sf \leadsto Sin^4 \ \theta +Cos^4 \ \theta =(Sin^2 \ \theta +Cos^2 \ \theta )^2- 2\ Sin^2 \ \theta \ Cos^2 \ \theta$

$\boxed{\bold{\gray{\large{Sin^2 \ \theta +Cos^2 \ \theta=1}}}}$

$\sf \leadsto Sin^4 \ \theta +Cos^4 \ \theta =(1)^2- 2 \ Sin^2 \ \theta \ Cos^2 \ \theta$

$\sf \leadsto Sin^4 \ \theta +Cos^4 \ \theta =1-2 \ Sin^2 \ \theta \ Cos^2 \ \theta$

$\sf Subtract \ by \ sin^2 \ \theta \ cos^2 \ \theta$

$\sf \leadsto Sin^4 \ \theta -Sin^2 \ \theta \ Cos^2 \ \theta+Cos^4 \ \theta = 1-3 \ Sin^2 \ \theta \ Cos^2 \ \theta$

$\sf Sin \ \theta + Cos \ \theta = m$

Square on both sides.

$\boxed{\bold{\large{\gray{(A + B)^2 = A^2 +2AB +B^2}}}}$

$\sf Sin^2 \ \theta + Cos^2 \ \theta+2\ sin \ \theta\ cos \ \theta = m^2$

$\boxed{\bold{\gray{\large{Sin^2 \ \theta +Cos^2 \ \theta=1}}}}$

$\sf 1+2\ sin \ \theta\ cos \ \theta = m^2$

$\sf 2\ sin \ \theta\ cos \ \theta = m^2-1$

$\sf sin \ \theta\ cos \ \theta = \dfrac{m^2-1}{2}$

Square on both sides

$\sf sin^2 \ \theta\ cos^2\ \theta = \dfrac{(m^2-1)^2}{4}$

Substitute the value of Sin² θ + Cos² θ

$\sf 1-3 \dfrac{(m^2-1)^2}{4}$

$\sf \dfrac{4}{4} -3 \dfrac{(m^2-1)^2}{4}$

$\sf \dfrac{4-3(m^2-1)^2}{4}$

$\sf Sin^6 \ \theta + Cos^6 \ \theta = \dfrac{3-4(m^2-1)^2}{4}$

HENCE PROVED.