Question

Cos 10a + cos 8a + 3cos 2a + 3cos 4a = 8cos a cos^3 3a

Answer 1

$\text{Consider,}$

$Cos\,10A + cos\,8A + 3\,cos\,2A+ 3\,cos\,4A$

$=Cos\,10A + cos\,8A + 3[cos\,2A+cos\,4A]$

$\text{Using,}$

$\boxed{\bf\;cosC+cosD=2\,cos(\frac{C+D}{2})\,cos(\frac{C-D}{2})}$

$\text{we get}$

$=2\,cos(\frac{10A+8A}{2})}\,cos(\frac{10A-8A}{2})+ 3[2\,cos(\frac{2A+4A}{2})\,cos(\frac{2A-4A}{2})]$

$=2\,cos\,9A\,cos\,A+ 3[2\,cos\,3A\;cos(-A)]$

$=2\,cos\,3(3A)\,cos\,A+6\,cos\,3A\;cosA$

$\text{Using}$

$\boxed{\bf\;cos\,3A=4\,cos^3A-3\,cosA}$

$=2[4\,cos^33A-3cos\,3A]cos\,A+6\,cos\,3A\;cosA$

$=8\,cos^33A\,cos\,A-6\,cos\,3A\,cos\,A+6\,cos\,3A\;cosA$

$=8\,cos^33A\,cos\,A$

$\therefore\textbf{cos\,10A + cos\,8A + 3\,cos\,2A+ 3\,cos\,4A= \bf\,8\,cos^33A\,cos\,A }$

Find more:

SinA+sin3A/cosA+cos 3A=tan2A prove​

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Answer 2

Answer:

Step-by-step explanation: