Question : If $\sin heta+\sin^{2} heta=1$, then the value of $\cos^{12} heta+3\cos^{10} heta+3\cos^{8} heta+\cos^{6} heta-1$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 0

Question

Question : If $\sin heta+\sin^{2} heta=1$, then the value of $\cos^{12} heta+3\cos^{10} heta+3\cos^{8} heta+\cos^{6} heta-1$ is:

Option 1: 1

Option 2: 2

Option 3: 3

Option 4: 0

Team Careers360 · Accepted Answer

Correct Answer: 0

Solution : Given: $\sin heta+\sin^{2} heta=1$
⇒ $\sin heta=1-\sin^{2} heta$
⇒ $\sin heta=\cos^2 heta$
Squaring both sides, we get,
⇒ $\sin^2 heta=\cos^4 heta$
⇒ $1-\cos^2 heta=\cos^4 heta$
⇒ $1=\cos^4 heta+\cos^2 heta$
Now, cubing both sides, we get,
⇒ $1^3=(\cos^4 heta+\cos^2 heta)^3$
By using $(a + b)^3 = a^3 + 3a^2b + 3ab^2 +b^3$
⇒ $1=\cos^{12} heta+3\cos^{10} heta+3\cos^{8} heta+\cos^{6} heta$
⇒ $\cos^{12} heta+3\cos^{10} heta+3\cos^{8} heta+\cos^{6} heta-1=0$
Hence, the correct answer is 0.

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Question : If sin⁡θ+sin2⁡θ=1, then the value of cos12⁡θ+3cos10⁡θ+3cos8⁡θ+cos6⁡θ−1 is:Option 1: 1Option 2: 2Option 3: 3Option 4: 0

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Question : If $\sin θ + \sin^{2} θ = 1$ , then the value of $\cos^{12} θ + 3 \cos^{10} θ + 3 \cos^{8} θ + \cos^{6} θ - 1$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 0