Question

what is the correct ans​

Answer 1

Given :-

♦ a = sin (π/4)

♦ b = cos (π/4)

♦ c = -cosec (π/4)

To find :-

♦ The value of a³+b³+c³

Solution :-

Given that

♦ a = sin (π/4)

♦ b = cos (π/4)

♦ c = -cosec (π/4)

We know that

♦ π = 180°

π/4 = 180°/4 = 45°

Therefore, π/4 = 45°

a = sin (π/4)

=> a = sin 45°

=> a = 1/√2

b = cos (π/4)

=> b = cos 45°

=> b = 1/√2

and

c = - cosec (π/4)

=> c = - cosec 45°

=> c = - √2

Now,

The value of a³ + b³ + c³

=> (1/√2)³+(1/√2)³+(-√2)³

=> [1/√(2×2×2)]+[1/√(2×2×2)]+[-√(2×2×2)]

=> [1/(2√2)] + [1/(2√2)] + (-2√2)

=> [2/(2√2)] -2√2

=> (1/√2)-2√2

=> [1-2√(2×2)]/√2

=> [1-(2×2)]/√2

=> (1-4)/√2

=> -3/√2

=> (-3×√2)/(√2×√2)

=> -3√2/2

Alternative Method:-

Given that

♦ a = sin (π/4)

♦ b = cos (π/4)

♦ c = -cosec (π/4)

They can be written as

a = sin 45° = 1/√2

b = cos 45° = 1/√2

c = - cosec 45° = -√2

Now,

a+b+c = (1/√2)+(1/√2)-(√2)

=> a+b+c = (2/√2)-(√2)

=> a+b+c = [2-√(2×2)]/√2

=> a+b+c = (2-2)/√2

=> a+b+c = 0/√2

=> a+b+c = 0

We know that

If a+b+c = 0 then a³+b³+c³ = 3abc

=> 3×sin 45° × cos 45° × - cosec 45°

=> 3 × (1/√2)×(1/√2)×(-√2)

=> -3√2/(√2×2)

=> -3√2/2

Therefore, a³+b³+c³ = -3√2/2

Answer :-

♦ The value of a³+b³+c³ is -3√2/2

Used formulae:-

♦ π = 180°

♦ sin 45° = 1/√2

♦ cos 45° = 1/√2

♦ cosec 45° = √2

♦ If a+b+c = 0 then a³+b³+c³ = 3abc