Question

$(sin3\theta +sin\theta ).sin\theta +(cos3\theta -cos\theta ).cos\theta =$

Answer 1

$(sin3\theta +sin\theta ).sin\theta +(cos3\theta -cos\theta )cos\theta$

$=sin3\theta sin\theta +si{n}^2\theta +cos3\theta cos\theta -co{s}^2\theta$

$=sin3\theta sin\theta +cos3\theta cos\theta +si{n}^2\theta -co{s}^2\theta$

$=cos(3\theta -\theta )+(-co{s}^2\theta )+si{n}^2\theta .$ $[\because cos({\theta }_1-{\theta }_2)=cos{\theta }_1,cos{\theta }_2+sin{\theta }_{1}sin{\theta }_2]$

$=cos2\theta -(co{s}^2\theta -si{n}^2\theta )$

$=cos2\theta -cos2\theta$ $(co{s}^2\theta -si{n}^2\theta =cos2\theta )$

$=0$