Determine if the following functions are even, odd, or neither of them.
f:x\longmapsto\sin\left(2x\right)-\cos\left(x\right)^2.\sin\left(x\right)
We have:
f\left(x\right)=\sin\left(2x\right)-\cos\left(x\right)^2.\sin\left(x\right)
Therefore:
f\left(-x\right)=\sin\left(-2x\right)-\cos\left(-x\right)^2.\sin\left(-x\right)
We know that:
- \sin\left(x\right) is odd.
- \cos\left(x\right) is even.
Thus:
f\left(-x\right)=-\sin\left(2x\right)-\cos\left(x\right)^2.\left[-\sin\left(x\right)\right]\\=-\sin\left(2x\right)+\cos\left(x\right)^2.\sin\left(x\right)\\=-\left(\sin\left(2x\right)-\cos\left(x\right)^2.\sin\left(x\right)\right)\\=-f\left(x\right)
This function is odd.
f:x\longmapsto \sin^2\left(2x\right)-\cos^2\left(x\right)
We have:
f\left(x\right)=\sin^2\left(2x\right)-\cos^2\left(x\right)
Therefore:
f\left(-x\right)=\sin^2\left(-2x\right)-\cos^2\left(-x\right)
We know that:
- \sin\left(2x\right) is odd.
- \cos\left(x\right) is even.
Thus:
f\left(-x\right)=\left[-\sin\left(2x\right)\right]^2-\left[\cos\left(x\right)^2\right]\\=\sin^2\left(2x\right)-\cos^2\left(x\right)\\=f\left(x\right)
This function is even.
f:x\longmapsto \tan\left(x\right)+\cot\left(x\right)
We have:
f\left(x\right)=\tan\left(x\right)+\cot\left(x\right)
Therefore:
f\left(-x\right)=\tan\left(-x\right)+\cot\left(-x\right)
We know that:
- \tan\left(x\right) is odd.
- \cot\left(x\right) is odd.
Thus:
f\left(-x\right)= -\tan\left(x\right) - \cot\left(x\right)\\=-\left(\tan\left(x\right)+\cot\left(x\right)\right)\\=-f\left(x\right)
This function is odd.
f:x\longmapsto\cos\left(x\right)\sin^2\left(x\right)+\cos^3\left(x\right).\sin^4\left(x\right)
We have:
f\left(x\right)=\cos\left(x\right)\sin^2\left(x\right)+\cos^3\left(x\right).\sin^4\left(x\right)
Therefore:
f\left(-x\right)=\cos\left(-x\right)\left[\sin\left(-x\right)\right]^2+\left[\cos\left(-x\right)\right]^3.\left[\sin\left(-x\right)\right]^4
We know that:
- \sin\left(x\right) is odd.
- \cos\left(x\right) is even.
Thus:
f\left(-x\right)=\cos\left(x\right)\left[-\sin\left(x\right)\right]^2+\cos\left(x\right)^3.\left[-\sin\left(x\right)\right]^4\\=\cos\left(x\right)\sin^2\left(x\right)+\cos^3\left(x\right).\sin^4\left(x\right)\\=f\left(x\right)
This function is even.
f\left(x\right) = \sin\left(x\right)+ \cos\left(x\right)
We have:
f\left(x\right)=\sin\left(x\right)+\cos\left(x\right)
Therefore:
f\left(-x\right)=\sin\left(-x\right)+\cos\left(-x\right)
We know that:
- \sin\left(x\right) is odd.
- \cos\left(x\right) is even.
Thus:
f\left(-x\right)= -\sin\left(x\right) +\cos\left(x\right)
And:
- f\left(-x\right) \ne f\left(x\right)
- f\left(-x\right) \ne -f\left(x\right)
This function is neither odd nor even.
f\left(x\right) = \sin\left(x\right)\left[1-\cos\left(x\right)\right]
We have:
f\left(x\right) = \sin\left(x\right)\left[1-\cos\left(x\right)\right]
Therefore:
f\left(-x\right) = \sin\left(-x\right)\left[1-\cos\left(-x\right)\right]
We know that:
- \sin\left(x\right) is odd.
- \cos\left(x\right) is even.
Thus:
f\left(-x\right) = -\sin\left(x\right)\left[1-\cos\left(x\right)\right]\\=-f\left(x\right)
This function is odd.
f\left(x\right)=\cos\left(x\right)\sin\left(x-\pi\right)+\cos\left(2x\right)\sin\left(x\right)
We have:
f\left(x\right)=\cos\left(x\right)\sin\left(x-\pi\right)+\cos\left(2x\right)\sin\left(x\right)
Therefore:
f\left(-x\right)=\cos\left(-x\right)\sin\left(-x-\pi\right)+\cos\left(-2x\right)\sin\left(-x\right)
We know that:
- \sin\left(x\right) is odd.
- \cos\left(x\right) is even.
Thus:
f\left(-x\right)=\cos\left(x\right)\left[-\sin\left(x+\pi\right)\right]+\cos\left(2x\right)\left[-\sin\left(x\right)\right]\\=- \cos\left(x\right)\sin\left(x+\pi\right)-\cos\left(2x\right)\sin\left(x\right)
For every angle \theta, we have:
\sin\left(\theta\right) = \sin\left(\theta + 2\pi\right)
Hence:
\sin\left(x-\pi\right) = \sin\left(x-\pi +2\pi\right) = \sin\left(x+\pi\right)
Therefore:
f\left(-x\right)=-\cos\left(x\right)\sin\left(x-\pi\right)-\cos\left(2x\right)\sin\left(x\right)\\=-f\left(x\right)
This function is odd.