2
$\begingroup$

Show that

$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$

Starting from the left hand side (LHS)

\begin{align} \text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\ &=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\ &=\cos^4A-\cos^2A\sin^2A+\sin^4A \end{align}

Can anyone help me to continue from here

$\endgroup$
3
$\begingroup$

Let $a=\cos^2 (A),b=\sin^2 (A) $ . Now use $a^3+b^3=(a+b)^3-3ab (a+b) $ also note that $a+b=1$. Hence the proof.

$\endgroup$
2
$\begingroup$

HINT:

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$ can also be written as

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

$$=\cos^4A-\cos^2A\sin^2A+\sin^4A$$ $$=\cos^4A-2\cos^2A\sin^2A+\sin^4A + \cos^2a\sin^2a$$ $$=(\cos^2A-\sin^2A)^2 + \cos^2A\sin^2A=1-3\sin^2A\cos^2A$$ $$(\cos^2A-\sin^2A)^2 + 4\cos^2A\sin^2A=1$$ $2\cos x\sin x=\sin2x$ -> $\sin^22x=4\cos^2\sin^2x$

and

$\cos^2x-\sin^2x =\cos2x$

$$\cos^22A + sin^22A=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.