Question

show that sin raised to 4 theta minus Cos raise to 4 theta is equal to 1 - 2 cos square theta​

Answer 1

Answer:

sin4x - cos4x = (1 - 2cos2x)

Step-by-step explanation:

sin4x - cos4x

= (sin2x)^2 - (cos2x)^2

= (sin2x + cos2x)(sin2x - cos2x)

= 1(sin2x - cos2x)         [Since sin2x + cos2x = 1]

= ((1 - cos2x) - cos2x)   [Since sin2x = 1 - cos2x]

= (1 - 2cos2x)

Thus, proved.

Answer 2

$\sin^4 \theta-\cos^4 \theta=1-2\cos^2 \theta$, proved.

Step-by-step explanation:

To prove that, $\sin^4 \theta-\cos^4 \theta=1-2\cos^2 \theta.$

L.H.S. $=\sin^4 \theta-\cos^4 \theta$

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

Using the formula,

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

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

Using trigonometric identity,

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

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

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

$=1-\cos^2 \theta-\cos^2 \theta$

Using trigonometric identity,

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

$=1-2\cos^2 \theta$

= R.H.S., proved.

Hence, $\sin^4 \theta-\cos^4 \theta=1-2\cos^2 \theta$, proved.