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(Solved): The three versions of the double-angle formula for cosine are given below. \[ \begin{array}{c} \cos ...
The three versions of the double-angle formula for cosine are given below. \[ \begin{array}{c} \cos (2 x)=2 \cos ^{2}(x)-1 \\ \cos (2 x)=1-2 \sin ^{2}(x) \\ \cos (2 x)=\cos ^{2}(x)-\sin ^{2}(x) \end{array} \] Under each set of assumptions, determine \( \cos (2 x) \). Part A: \( \sin (x)=\frac{3}{4} \) and \( \cos (x)>0 \) \[ \cos (2 x)= \] Part B: \( \cos (x)=-\frac{1}{6} \) and \( \sin (x)>0 \) \[ \cos (2 x)= \] \( \operatorname{Part} C: \sin (x)=-\frac{2}{9} \) and \( \cos (x)<0 \) \[ \cos (2 x)= \] Determine the value of \( \cos ^{2}(2 x)+\sin ^{2}(2 x) \). \[ \cos ^{2}(2 x)+\sin ^{2}(2 x)= \]
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Solution :- A) => We have sin(x) = 3/4 & cos?(x)=1?sin2(x)=1?916=716 cos2(x)=716 Now cos?(2x)=2cos2(x)?1=2×716?1=?18(1)cos?(2x)=1?2sin2(x)=1?2×916=1?9
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