Question

If 'θ''θ' is in the III quadrant then 4sin4θ+sin22θ−−−−−−−−−−−−√+4cos2(π4−θ2)=​

Answer 1

Answer:

here is your ans

Step-by-step explanation:

√{4sin

4

θ+sin

2

2θ}=√{4sin

2

θ(sin

2

θ+cos

2

θ)}

(4sin

2

θ)

=2∣sinθ∣.

But since θ lies in the third quadrant, we have sinθ<0 and ∣sinθ∣=−sinθ.

Hence √4sin

4

θ+sin

2

2θ=−2sinθ.

and 4cos

2

(

4

π

−

2

θ

)=2[1+cos(

2

π

−θ)]=2+2sinθ.

Hence the given expression

=−2sinθ+2+2sinθ=2.

[Note that ∣x∣=x if x≥0 and =−x if x≤0

Answer 2

Answer:

-2sin0+2+2sin0=2

Step-by-step explanation:

hope it helps