Question

plz answer this guys​

Answer 1

Answer:

2 a₆ - 3 a₄ = - 1

Step-by-step explanation:

Given----> aₙ = Cosⁿα + Sinⁿα

To find ----> Value of ( 2 a₆ - 3 a₄ )

Concept used ---->

1)

$( \: {a}^{3} \: + \: {b}^{3} \: ) \: = ( \: a \: + \: b \: ) \: ( \: {a}^{2} \: + {b}^{2} \: - \: ab \: )$

2)

$( \: a \: + \: b \: ) ^{2} \: = \: {a}^{2} \: + \: {b}^{2} \: + \: 2ab$

3)

${sin}^{2} \alpha \: + \: \: {cos}^{2} \alpha = \: 1$

Solution-----> ATQ,

aₙ = Cosⁿα + Sinⁿα

a₆ = Cos⁶α + Sin⁶α

= ( Cos²α )³ + ( Sin²α )³

= ( Cos²α + Sin²α ) { ( Cos²α )² + ( Sin²α )² - Cos²α Sin²α }

= ( 1 ) { ( Cos²α )² + ( Sin²α )² + 2 Sin²α Cos²α - 3 Sin²α Cos²α }

= ( Cos²α + Sin²α )² - 3 Sin²α Cos²α

= ( 1 )² - 3 Sin²α Cos²α

= 1 - 3 Sin²α Cos²α

Now ,

a₄ = Cos⁴α + Sin⁴α

= ( Cos²α )² + ( Sin²α )² + 2 Sin²α Cos²α

- 2 Sin²α Cos²α

= ( Cos²α + Sin²α )² - 2 Sin²α Cos²α

= ( 1 )² - 2 Sin²α Cos²α

= 1 - 2 Sin²α Cos²α

Now, 2 a₆ - 3 a₄

$=2 \: ( \: 1 \: - 3 \ {sin}^{2} \alpha \: {cos}^{2} \alpha \: ) \: - 3 \: ( \: 1 - 2 \: {sin}^{2} \alpha \: {cos}^{2} \alpha )$

$= 2 \: - \: 6 {sin}^{2} \alpha \: {cos}^{2} \alpha \: - 3 \: + \: 6 \: {sin}^{2} \alpha \: {cos}^{2} \alpha$

$=2 \: - \: 3$

$= - \: 1$

Answer 2

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