Question

Please answer this 11th Trignometry ques ​

Answer 1

Answer:

( B ).

Step-by-step explanation:

Given expression is :  

 $3\bigg[\sin^4\bigg(\dfrac{3\pi}{2}-\alpha\bigg)+\sin^4(3\pi+\alpha)\bigg]-2\bigg[\sin^6\bigg(\dfrac{\pi}{2}+\alpha\bigg)+\sin^6(5\pi-\alpha)\bigg]$

   We know sine and cosine are periodic at 2π.

Therefore, ( now : solving in terms )

$\sin^4\bigg(\dfrac{3\pi}{2}-\alpha\bigg) = \sin^4\bigg(-\dfrac{\pi}{2}-\alpha\bigg) = \cos^4\alpha$

$\sin^4(3\pi+\alpha)=\sin^4(\pi+\alpha)=\sin^4\alpha$

$\sin^6\bigg(\dfrac{\pi}{2}+\alpha\bigg)=\cos^6\alpha$

$\sin^6(5\pi-\alpha)=\sin^6(\pi-\alpha)=\sin^6\alpha$

  Continued : After solving( terms ) above expression looks like :

⇒ 3[ cos⁴α + sin⁴α ] - 2[ cos⁶α + sin⁶α ]

       Expanding the term with 3 :

⇒ [ ( cos⁴α + sin⁴α ) + 2( cos⁴α + sin⁴α ) ] - 2[ cos⁶α + sin⁶α ]

⇒ ( cos⁴α + sin⁴α ) + 2[ cos⁴α + sin⁴ - cos⁶α - sin⁶α ]

⇒ ( cos⁴α + sin⁴α ) + 2[ cos⁴α - cos⁶α + sin⁴α - sin⁶α ]

⇒ ( cos⁴α + sin⁴α ) + 2[ cos⁴α( 1 - cos²α ) + sin⁴α( 1 - sin²α ) ]

⇒ ( cos⁴α + sin⁴α ) + 2[ cos⁴α.sin²α + sin⁴α.cos²α ]

⇒ ( cos⁴α + sin⁴α ) + 2cos²α.sin²α( cos²α + sin²α )

⇒ ( cos⁴α + sin⁴α ) + 2cos²α.sin²α( 1 )             { cos²A + sin²A = 1 }

⇒ cos⁴α + sin⁴α + 2cos²α.sin²α

 Using a⁴ + b⁴ + 2a²b² = ( a² + b² )²

⇒ ( cos²α + sin²α )²

⇒ ( 1 )²

⇒  1

      Hence the required value of the given expression is 1.

Answer 2

Question :-

$3( { \sin( \frac{3\pi}{2} - \alpha ) }^{4} + { \sin(3\pi + \alpha ) }^{4} )- 2( { \sin( \frac{\pi}{2} + \alpha ) }^{6} + { \sin(5\pi - \alpha ) }^{6} )$

Solution :-

We have two brackets over here . We'll solve each one by one . So let's go through first one .

$3 [ {\sin ( \frac{3\pi}{2} - \alpha) }^{4} + {\sin(3\pi + \alpha ) }^{4}]$

Now as we know that:-

As 3π/2 = 3(180/2)

= 3(90) = 270°

↭ Sin (90 - A ) = Cos A

( because all angles are positive in 1st quadrant )

↭ Sin ( 180 + A ) = - Sin A

3π = 540° .

As 3 π is odd value and we know odd valve of π is equal to 180°

( because sine is negative in 3rd quadrant )

And 3π/2 = 3(90)° , 3π = 3(180)°

So using these identities .

${\sin (\frac {3\pi}{2} - \alpha )}^{4} = { (- \cos ( \alpha)) }^{4}$

${\sin ( 3\pi + \alpha )}^{4} = { (- \sin ( \alpha))}^{4}$

↭ Because both have even power then there negative sign will convert into positive sign .

$3( { \cos( \alpha ) }^{4} + { \sin( \alpha ) }^{4} )$

This is equation 1st .

Now taking the second one .

$- 2( { \sin( \frac{\pi}{2} + \alpha ) }^{6} + { \sin(5\pi - \alpha ) }^{6} )$

Now as we know that :-

↭ Sin ( 90 + A) = - Cos (A)

( because cos is negative in second quadrant )

↭ Sin ( 180 - A ) = Sin A

( because sine is positive in third quadrant )

So using these identities :-

${ \sin( \frac{\pi}{2} + \alpha ) }^{6} = {\cos (\alpha)}^{6}$

( because the power is even negative sign converted into positive one )

${ \sin(5\pi - \alpha ) }^{6} = { \sin( \alpha ) }^{6}$

$-2[ {\cos ( \alpha)}^{6} + {\sin \alpha}^{6}]$

This is equation 2nd

Now adding both equations :-

$3( { \cos( \alpha ) }^{4} + { \sin( \alpha ) }^{4} )-2[ {\cos ( \alpha)}^{6} + {\sin \alpha}^{6}]$

Some more identities :-

↭ a³ + b³ = ( a + b ) ( a² - ab +b²)

↭ a² + b² = ( a+b)² - 2ab .

Here A = alpha .

3 [ ( cos² A)² + ( sin ²A)² ] - 2 [ ( cos²A)³ +( sin²A)³]

Applying the above mentioned identities .

3 [ ( cos²A + sin²A )² - 2 cos²A.sin²A ] -2 { ( cos²A+sin²A) [ ( cos²A)² - cos²A .sin²A +( sin²b)²] }

Sin²A + Cos²A = 1 .

3 [ (1)² - 2 cos²Asin²A ] -2 [ (1) (( cos²A )²+ (sin²A )² - cos²A . sin²A ]

Again implying a² + b² here

( 3 - 6 cos² A sin²A ) - 2 [ (cos²A + sin²A)² - 2cos²A. sin²A - cos²A . sin²A ]

( 3 - 6cos²A . sin²A ) - 2 [ 1 - 3 cos²A . sin²A ]

3 - 6 cos²A .sin²A - 2 +6 cos²A. sin²A

↭ 3 - 2 = 1

↭ So 1 is the answer