Class 10 - Trigonometry Identity Based Question
Question: If sinπ + sin²π = 1, find the value of
cos¹²π + 3cos¹⁰ π + 3 cos⁸π + cos⁶π + 2 cos⁴π + 2 cos²π - 2
Answer:
Given,
sinπ + sin²π = 1
sinπ + sin²π = sin²π + cos²π
∴ sinπ = cos²π
= cos¹²π + 3cos¹⁰ π + 3 cos⁸π + cos⁶π + 2 cos⁴π + 2 cos²π - 2
= (cos¹²π + 3cos¹⁰ π + 3 cos⁸π + cos⁶π) + 2 (cos⁴π + cos²π - 1)
= [(cos⁴π)³ + 3(cos⁴π)²(cos²π) + 3(cos⁴π)(cos²π)² + (cos²π)³] + 2(sin²π + cos²π - 1)
= (cos⁴π + cos²π)³ + 2 (1 - 1)
= (sin²π + cos²π)³ + 2(0)
= (1)³ + 0
= 1
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