Sin^4x+cos^4x=1-1÷2sin^2x
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4
Sin⁴x+cos⁴x
→ (sin²x)²+(cos²)²
→ (sin²x)²+(cos²)² +2sin²xcos²x - 2sin²xcos²x
→ (sin²x+cos²)²-2sin²xcos²
→ 1-2sin²xcos²x
→1-4sin²xcos²x/2
→1-sin²2x/2
LHS=RHS
Answered by
3
Answer:
f(x)=sin²x+(1−sin²x)²
=1+sin²x−2sin²x+sin⁴x
=1−sin²x(1−sin²x)
=1−sin²xcos²x
sin²x∈[−1,1]
⇒sin²2x∈[0,1]
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