Question

If sin 0 + sin-o = 1 then cos12 e + 3 cos 10 e + 3 cos 8 e + cosó e + 2 cost 0 + 2cos2 0–2 = ?​

Answer 1

Step-by-step explanation:

${\Large{\mathbf{\underline{\orange{Proper \: Question :-}}}}}$

If sin θ + sin² θ = 1, find the value of :

cos¹²θ + 3 cos¹⁰θ + 3 cos⁸θ + cos⁶θ + 2 cos⁴θ + 2 cos²θ - 2

${\Large{\mathbf{\underline{\orange{Solution :-}}}}}$

Consider the given data :

➠ sin θ + sin² θ = 1

➠ sin θ = 1 - sin² θ

➠ sin θ = cos² θ

We have

➠ cos¹²θ + 3 cos¹⁰θ + 3 cos⁸θ + cos⁶θ + 2 cos⁴θ + 2 cos²θ - 2

➠ cos¹²θ + 3 cos¹⁰θ + 3 cos⁸θ + cos⁶θ + 2(cos⁴θ + cos²θ - 1)

➠ (cos⁴ θ + cos² θ)³ + 2(cos⁴θ + cos²θ - 1)

Since sin θ = cos² θ

Therefore

➠ (sin² θ + sin θ)³ + 2(sin² θ + sin² θ - 1)

➠ 1³ + 2(1 - 1)

➠ 1 + 0

➠ 1

${\Large{\mathfrak{\fbox{\pink{The required answer is 1}}}}}$

Answer 2

$\large \fbox{Correct Question:}$

If sin θ + sin² θ = 1, find the value of :

cos¹²θ + 3 cos¹⁰θ + 3 cos⁸θ + cos⁶θ + 2 cos⁴θ + 2 cos²θ - 2

$\large \fbox{Required Answer:}$

➣sin θ + sin² θ = 1

➣ sin θ = 1 - sin² θ

➣sin θ = cos² θ

We have

➣cos¹²θ + 3 cos¹⁰θ + 3 cos⁸θ + cos⁶θ + 2 cos⁴θ + 2 cos²θ - 2

➣cos¹²θ + 3 cos¹⁰θ + 3 cos⁸θ + cos⁶θ + 2(cos⁴θ + cos²θ - 1)

➣ (cos⁴ θ + cos² θ)³ + 2(cos⁴θ + cos²θ - 1)

Since sin θ = cos² θ

Therefore,

➣(sin² θ + sin θ)³ + 2(sin² θ + sin² θ - 1)

➣ 1³ + 2(1 - 1)

➣ 1 + 0

$➣ \: \underline{\boxed{\frak{1}}} \: \pink{ \bigstar}$

• Hence, the answer is 1.

Thank you!

@itzshivani