Question

A teacher writes two cubic identities on board. (i)(a+b)^3=a^3+b^3+3ab(a+b) (ii) (a-b)^3=a^3-b^3-3ab(a-b). Which identity will be used to prove sin^6+cos^6+3sin^2cos^2=1.Explain​

Answer 1

Answer:

(i)

Step-by-step explanation:

(sin^2)^3 + (cos^2)^3 + 3 sin^2 cos^2 (sin^2 + cos^2)

so here a = sin^2 and b = cos^2

sin^2 + cos^2 = 1 according to trigno formulas

so by given formula

(sin^2 + cos^2)^3 = (1)^3 = 1

Answer 2

$\sin^6 A +\cos^6 A+3\sin^2 A\cos^2 A=1$, proved.

Step-by-step explanation:

We have,

The  two cubic identities on board

(i) $(a+b)^3=a^3+b^3+3ab(a+b)$

(ii) $(a-b)^3=a^3-b^3-3ab(a-b)$

To prove that, $\sin^6 A+\cos^6 A+3\sin^2 A\cos^2 A=1$

∴ $(\sin^2 A+\cos^2 A)^{3}$

Using identity (1),

$(a+b)^3=a^3+b^3+3ab(a+b)$

$(\sin^2 A+\cos^2 A)^{3}=(\sin^2 A)^{3} +(\cos^2 A)^{3}+3\sin^2 A\cos^2 A(\sin^2 A+\cos^2 A)$

⇒ $(1)^{3}=(\sin^2 A)^{3} +(\cos^2 A)^{3}+3\sin^2 A\cos^2 A(1)$

[ ∵ $\sin^2 A+\cos^2 A=1$]

⇒ $1=(\sin^2 A)^{3} +(\cos^2 A)^{3}+3\sin^2 A\cos^2 A$

⇒ $1=\sin^6 A +\cos^6 A+3\sin^2 A\cos^2 A$, proved.

Hence,$\sin^6 A +\cos^6 A+3\sin^2 A\cos^2 A=1$, proved.