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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If sin θ = ^ 2 θ , then the value of ^ 12 θ + 3 ^ 10 θ + 3 ^ 8 θ +...

If sinθ=cos⁡2θsin \theta = \cos^{2} \theta, then the value of cos⁡12θ+3cos⁡10θ+3cos⁡8θ+cos⁡6θ−1\cos^{12} \theta + 3\cos^{10} \theta + 3\cos^{8} \theta + \cos^{6} \theta - 1 is equal to

Correct answer

0

Step-by-step solution

sinθ=cos⁡2θ⇒sinθ+sin⁡2θ=1sin \theta = \cos^{2} \theta ⇒ sin \theta + \sin^{2} \theta = 1
Now, cos⁡6θ(cos⁡6θ+3cos⁡4θ+3cos⁡2θ+1)−1\cos^{6} \theta\left(\cos^{6} \theta + 3\cos^{4} \theta + 3\cos^{2} \theta + 1\right) - 1
=cos⁡6θ(cos⁡2θ+1)3−1= \cos^{6} \theta{\left(\cos^{2} \theta + 1\right)}^{3} - 1
=(cos⁡2θ)3(sinθ+1)3−1= {\left(\cos^{2} \theta\right)}^{3}{\left(sin \theta + 1\right)}^{3} - 1
=sin⁡3θ(sinθ+1)3−1= \sin^{3} \theta{\left(sin \theta + 1\right)}^{3} - 1
=(sin⁡2θ+sinθ)3−1=1−1=0= {\left(\sin^{2} \theta + sin \theta\right)}^{3} - 1 = 1 - 1 = 0
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