Math, asked by stuthiishana9416, 1 year ago

if cos theta + cos squared theta equal to 1 then prove that sin to the power 12 theta + 3 sin to the power 10 theta + 3 sin to the power 8 theta + sin to the power 6 theta + 2 Sin to the power 4 theta + 2 Sin to the power 4 theta minus 2 equal to 1

Answers

Answered by ShuchiRecites
100

Given

→ cos∅ + cos²∅ = 1

→ cos∅ + 1 - sin²∅ = 1

→ cos∅ = sin²∅

Solution

→ sin¹²∅ + 3sin^(10)∅ + 3 sin^8∅ + sin^6∅ + 2sin⁴∅ + 2 sin²∅ - 2 = 1

→ (sin⁴∅)³ + 3 (sin⁴∅)² (sin²∅) + 3 sin⁴∅(sin²∅)² + (sin²∅)³ + 2(sin²∅)² + 2 sin²∅ - 2 = 1

→ (sin⁴∅ + sin²∅)³ 2(sin²∅)² + 2 sin²∅ - 2 = 1

→ [(sin²∅)² + sin²∅]³ + 2(sin²∅)² + 2 sin²∅ - 2 = 1

→ (cos²∅ + cos∅)³ + 2(cos²∅ + cos∅) - 2 = 1

Since cos²∅ + cos∅ = 1

→ 1³ + 2(1) - 2 = 1

→ 1 = 1

L.H.S = R.H.S

Hence Proved

Answered by BrainlyWriter
79

 \bold {\huge {Your ~answer :-}}

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EXPLAINATION ➣

Given-\\</p><p>cos\theta+cos^2\theta=1\\To\:prove-\\</p><p>sin^{12}\theta+3sin^{10}\theta+3sin^8\theta+sin^6\theta+2sin^4\theta+2sin^2\theta-2 =1

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

→cos θ+cos²θ=1

⇒cos θ=1−cos²θ

⇒cos θ= sin²θ

Taking LHS,

⇒(sin⁴θ)3+3 (sin⁴θ)2(sin²θ) +3 (sin⁴θ) (sin²θ)2+(sin²θ)3+2(sin⁴θ+sin²θ)−2(sin⁴θ+sin²θ)3+2(sin⁴θ+sin²θ)−2       

   ➣ Note--- [(a+b)³=a³+b³+3a²b+3ab²]

⇒[(sin²θ)2+sin²θ]3+2[(sin²θ)2+sin²θ]−2   

 [ sin²θ= cosθ] given

⇒(cos²θ+sin²θ)3+2(cos²θ+sin²θ)−2            

[sin²θ+cos²θ = 1]

⇒1³+2−2

⇒1 = 1

LHS=RHS Hence Proved.

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