[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)Prove:
1 answer:
Given:

To prove:
The given statement.
Proof:
We have,



![[\because a^2+b^2=(a+b)^2-2ab]](https://tex.z-dn.net/?f=%5B%5Cbecause%20a%5E2%2Bb%5E2%3D%28a%2Bb%29%5E2-2ab%5D)
![[\because \cos^2 \alpha+\sin^2\alpha=1]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Ccos%5E2%20%5Calpha%2B%5Csin%5E2%5Calpha%3D1%5D)

Now,

![[\because \cos 2\theta=2\cos^2\theta -1]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Ccos%202%5Ctheta%3D2%5Ccos%5E2%5Ctheta%20-1%5D)
![RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]](https://tex.z-dn.net/?f=RHS%3D%5Cdfrac%7B1%7D%7B4%7D%5B2%2B2%5Ccos%5E2%202%5Calpha%5D)
![[\because \cos 2\theta=2\cos^2\theta -1]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Ccos%202%5Ctheta%3D2%5Ccos%5E2%5Ctheta%20-1%5D)
![[\because (a-b)^2=a^2-2ab+b^2]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5D)
![RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]](https://tex.z-dn.net/?f=RHS%3D%5Cdfrac%7B1%7D%7B4%7D%5B2%2B8%5Ccos%5E4%20%5Calpha-8%5Ccos%20%5Calpha%2B2%5D)
![RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]](https://tex.z-dn.net/?f=RHS%3D%5Cdfrac%7B1%7D%7B4%7D%5B4%2B8%5Ccos%5E4%20%5Calpha-8%5Ccos%20%5Calpha%5D)



Hence proved.
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