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
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
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EXPLAINATION ➣
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→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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