Question

the value of (cos4A- sin4A) is equal to
(a) 1-2cos
(b) 2 sinA-1
(c) sin2A-cos2a
(d) 2cos2A-1​

Answer 1

Answer:

The value of \cos ^ { 4 } A - \sin ^ { 4 } A is equal to 2 \cos^2 A - 1

Solution:

Given,

\cos ^ { 4 } A - \sin ^ { 4 } A

The above equation is expressed as

\begin{array} { l } { \cos ^ { 4 } A = \left( \cos ^ { 2 } A \right) ^ { 2 } } \\\\ { \sin ^ { 4 } A = \left( \sin ^ { 2 } A \right) ^ { 2 } } \\\\ { \cos ^ { 4 } A - \sin ^ { 4 } A = \left( \cos ^ { 2 } A \right) ^ { 2 } - \left( \sin ^ { 2 } A \right) ^ { 2 } } \end{array}\\

The above equation is in the form of a^2-b^2=(a+b)(a-b)

\left( \cos ^ { 2 } A \right) ^ { 2 } - \left( \sin ^ { 2 } A \right) ^ { 2 } = \left( \cos ^ { 2 } A + \sin ^ { 2 } A \right) \left( \cos ^ { 2 } A - \sin ^ { 2 } A \right)

We know that  \left( \cos ^ { 2 } A + \sin ^ { 2 } A = 1 \right)

Therefore,

\begin{array} { r } { \cos ^ { 4 } A - \sin ^ { 4 } A = ( 1 ) \times \left( \cos ^ { 2 } A - \left( 1 - \cos ^ { 2 } A \right) \right. }\\ \\ { = \cos ^ { 2 } A - 1 + \cos ^ { 2 } A } \\\\ { = 2 \cos ^ { 2 } A - 1 } \end{array}

Therefore,

\cos ^ { 4 } A - \sin ^ { 4 } A = 2 \cos ^ { 2 } A - 1

Answer 2

The required "option d) $2\cos^2A-1$" is correct.

Step-by-step explanation:

We have,

$\cos^4A- \sin^4A$

To find, the value of $\cos^4A- \sin^4A$) is equal to:

∴ $\cos^4A- \sin^4A$

= $(\cos^2A)^2- (\sin^2A)^2$

Using the algebraic identity,

$a^{2} -b^{2}$ = (a + b)(a - b)

= $(\cos^2A+\sin^2A)(\cos^2A-\sin^2A)$

Using the trigonometric identity,

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

= $(1)(\cos^2A-\sin^2A)$

= $\cos^2A-\sin^2A$

Using the trigonometric identity,

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

⇒ $\sin^2\theta$ = 1 - $\cos^2\theta$

= $\cos^2A-(1-\cos^2A)$

= $\cos^2A-1+\cos^2A$

= $2\cos^2A-1$

∴ $\cos^4A- \sin^4A$ = $2\cos^2A-1$

Thus, the required "option d) $2\cos^2A-1$" is correct.