Question

sin 8 theta cos theta - sin 6 theta cos 3 theta/ cos 2 theta cos theta - sin 3 theta sin 4 theta

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\dfrac{sin\,8\theta\;cos\,\theta-sin\,6\theta\;cos\,3\theta}{cos\,2\theta\;cos\,\theta-sin\,3\theta\;sin\,4\theta}}$

$\underline{\textbf{To simplify:}}$

$\mathsf{\dfrac{sin\,8\theta\;cos\,\theta-sin\,6\theta\;cos\,3\theta}{cos\,2\theta\;cos\,\theta-sin\,3\theta\;sin\,4\theta}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{sin\,8\theta\;cos\,\theta-sin\,6\theta\;cos\,3\theta}{cos\,2\theta\;cos\,\theta-sin\,3\theta\;sin\,4\theta}}$

$\textsf{Multiply both numerator and denominator by 2}$

$\mathsf{=\dfrac{2\,sin\,8\theta\;cos\,\theta-2\,sin\,6\theta\;cos\,3\theta}{2\,cos\,2\theta\;cos\,\theta-2\,sin\,3\theta\;sin\,4\theta}}$

$\textsf{Using the following identities, we get}$

$\boxed{\begin{minipage}{6cm} \\2\,sin\,A\;cos\,B=sin(A+B)+sin(A-B)\\\\2\,cos\,A\;sin\,B=sin(A+B)-sin(A-B)\\\\2\,cos\,A\;cos\,B=cos(A+B)+cos(A-B)\\\\2\,sin\,A\;sin\,B=cos(A-B)-cos(A+B)\\ \end{minipage}}$

$\mathsf{=\dfrac{[sin(8\theta+\theta)+sin(8\theta-\theta)]-[sin(6\theta+3\theta)+sin(6\theta-3\theta)]}{[cos(2\theta+\theta)+cos(2\theta-\theta)]-[cos(3\theta-4\theta)-cos(3\theta+4\theta)]}}$

$\mathsf{=\dfrac{[sin\,9\theta+sin\,7\theta]-[sin\,9\theta+sin\,3\theta]}{[cos\,3\theta+cos\,\theta]-[cos\,\theta-cos\,7\theta]}}$

$\mathsf{=\dfrac{sin\,9\theta+sin\,7\theta-sin\,9\theta-sin\,3\theta}{cos\,3\theta+cos\,\theta-cos\,\theta+cos\,7\theta}}$

$\mathsf{=\dfrac{sin\,7\theta-sin\,3\theta}{cos\,3\theta+cos\,7\theta}}$

$\mathsf{=\dfrac{sin\,7\theta-sin\,3\theta}{cos\,3\theta+cos\,7\theta}}$

$\textsf{Using the following identities, we get}$

$\boxed{\begin{minipage}{8cm} \\sin\,C-sin\,D=2\,cos\left(\dfrac{C+D}{2}\right)\,sin\left(\dfrac{C-D}{2}\right)\\\\cos\,C+cos\,D=2\,cos\left(\dfrac{C+D}{2}\right)\,cos\left(\dfrac{C-D}{2}\right)\\ \end{minipage}}$

$\mathsf{=\dfrac{2\,cos\left(\dfrac{7\theta+3\theta}{2}\right)\,sin\left(\dfrac{7\theta-3\theta}{2}\right)}{2\,cos\left(\dfrac{3\theta+7\theta}{2}\right)\,cos\left(\dfrac{3\theta-7\theta}{2}\right)}}$

$\mathsf{=\dfrac{2\,cos\,5\theta\;sin\,2\theta}{2\,cos\,5\theta\,cos(-2\theta)}}$

$\mathsf{=\dfrac{2\,cos\,5\theta\;sin\,2\theta}{2\,cos\,5\theta\,cos\,2\theta}}$

$\mathsf{=\dfrac{sin\,2\theta}{cos\,2\theta}}$

$\mathsf{=tan\,2\theta}$

Answer 2

answer plz mark me as brain list