Question

1. If Sin 0 + Cos 0 = x, Prove that Sin⁴0 + Cos⁴0 = 2- (x2-1) / 2.
2. If Sin 0 + Cos 0 = a, Prove That Sin60 +Cos6 0 = 4-3 (a²- 1)² / 4.​

Answer 1

Given : sin ∅ + cos ∅ = x

On squaring both sides,

→ (sin ∅ + cos ∅)² = x²

→ sin²∅ + cos²∅ + 2 cos∅ sin∅ = x²

→ 1 + 2 sin∅·cos∅ = x²

→ 2 cos∅ · sin∅ = (x² - 1)

→ cos∅ · sin∅ = (x² - 1)/2

→ sin²∅ + cos²∅ = 1

→ (sin²∅ + cos²∅)² = 1

→ sin⁴∅ + cos⁴∅ + 2·sin²∅·cos²∅ = 1

→ sin⁴∅ + cos⁴∅ + 2(sin∅·cos∅)² = 1

→ sin⁴∅ + cos⁴∅ + 2[(x² - 1)²/4]= 1

→ sin⁴∅ + cos⁴∅ = 1 - (x² - 1)²/2

→ sin⁴∅ + cos⁴∅ = [2 - (x² - 1)²]/2

Hence Proved

______________________________

2. Given : sin∅ + cos∅ = a

On squaring both sides,

→ sin²∅ + cos²∅ + 2·cos∅·sin∅ = a²

→ 2·cos∅·sin∅ = a² - 1

→ cos∅·sin∅ = (a² - 1)/2

→ sin⁶∅ + cos⁶∅ = (sin²∅)³ + (cos²∅)³

→ (sin²∅ + cos²∅)³ - 3(sin∅·cos∅)²(sin²∅ + cos²∅)

→ 1³ - 3(a² - 1)²/4 × 1

→ [4 - 3(a² - 1)²]/4

Hence Proved

Answer 2

$\huge\boxed{Answer-1}$

According to the given questions squaring both sides.

sin ∅ + cos ∅² = x²

sin²∅ + cos²∅ + 2 cos∅ sin∅ = x²

1 + 2 sin∅ cos∅ = x²

2 cos∅ sin∅ = (x² - 1)

cos∅ sin∅ = (x² - 1)/2

sin²∅ + cos²∅ = 1

sin²∅ + cos²∅² = 1

sin⁴∅ + cos⁴∅ + 2sin²∅cos²∅ = 1

sin⁴∅ + cos⁴∅ + 2(sin∅cos∅)² = 1

sin⁴∅ + cos⁴∅ + 2[(x² - 1)²/4]= 1

sin⁴∅ + cos⁴∅ = 1 - (x² - 1)²/2

sin⁴∅ + cos⁴∅ = [2 - (x² - 1)²]/2

Proved that Sin⁴∅+ Cos⁴∅ = 2- (x2-1) / 2

$\huge\boxed{Answer-2}$

According to the given questions squaring both sides.

sin²∅ + cos²∅ + 2 cos∅ sin∅ = a²

2cos∅ sin∅ = a² - 1

cos∅sin∅ = (a² - 1)/2

sin⁶∅ + cos⁶∅ = (sin²∅)³ + (cos²∅)³

(sin²∅ + cos²∅)³ - 3(sin∅·cos∅)²

(sin²∅ + cos²∅)

1³ - 3(a² - 1)²/4 × 1

(4 - 3(a² - 1)²)/4

$(\frac{(4 - 3(a {}^{2} - 1 {}^{2} )}{4} {}^{ } )$

Proved that Sin60 +Cos6 0 = 4-3 (a²- 1)² / 4.