Question

show that sin^2 theta +sin^2 (theta+60)+sin^2 theta =3/2​

Answer 1

Answer:

$sin2θ+sin2(60∘+θ)+sin2(60∘−θ)=sin2θ+21(1−cos(120∘+2θ))+21(1−cos(120∘−2θ))</p><p></p><p>                                                                     =sin2θ+1−21(cos(120∘+2θ)+cos(120∘−2θ))</p><p></p><p></p><p>                                                                     =1+sin2θ−21(2cos120∘cos2θ)</p><p></p><p></p><p>                                                                     =1+sin2θ+21(1−2sin2θ)</p><p></p><p></p><p>                                                                     =1+21=23</p><p></p><p></p><p>$

Answer 2

answer is 15.3/.14 hope it's helpfullSince we're already given the length of the rectangle, 4.5 cm, and its width, 6 cm, then we can find the area A of the rectangle by simply substituting ...

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