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f(x)=1-2sin(x)^2cos(x)^2-2sin(x)^4-cos(2x)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

1−2⋅sin2(x)⋅cos2(x)−2⋅sin4(x)−cos(2x)1-2\cdot sin^2\left(x\right)\cdot cos^2\left(x\right)-2\cdot sin^4\left(x\right)-cos\left(2x\right)

Step-by-step Solution

1

Use the trigonometric identity: 1−cos⁡(2x)=2sin⁡(x)21-\cos\left(2x\right)=2\sin\left(x\right)^2

2sin⁡(x)2−2sin⁡(x)2cos⁡(x)2−2sin⁡(x)42\sin\left(x\right)^2-2\sin\left(x\right)^2\cos\left(x\right)^2-2\sin\left(x\right)^4

Final answer to the exercise

2sin⁡(x)2−2sin⁡(x)2cos⁡(x)2−2sin⁡(x)42\sin\left(x\right)^2-2\sin\left(x\right)^2\cos\left(x\right)^2-2\sin\left(x\right)^4

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1−2·sin2(x)·cos2(x)−2·sin4(x)−cos(2x)
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