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General Solution of cos theta = cos alpha

Identity tran...

Question

Identity transformations of Trigonometric Expressions.
Prove the following identities.
$1 - c o s (2 x - π) - c o s (4 x + π) + c o s (6 x - 2 π) = 4 c o s x \times c o s 2 x c o s 3 x .$

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Solution

$1 - c o s (2 x - π) - c o s (4 x + π) + c o s (6 x - 2 π) = 4 c o s x c o s 2 x . c o s x$
$= c o s (2 x - π) = c o s (π - 2 x) = - c o s 2 x$ ( $I I^{2 d}$ quadrant cosine -v)
$c o s (6 x - 2 π) = c o s (2 π - 6 x) = c o s 6 x$ ( $I V^{t h}$ quadrant cosine +ve)
$c o s (x + 4 x) = - c o s 4 x$ ( $I I I^{r d}$ quadrant cosine -ve)
So finally
$1 + c o s 2 x + c o s 4 x + c o s 6 x = 4 c o s x c o s 2 x c o s 3 x$
take RHS
RHS $= 4 c o s x c o s 2 x c o s 3 x$
Using $2 c o s A c o s B = c o s (A + B) + c o s (A - B)$
RHS $= 2 (2 c o s x c o s 2 x) c o s 3 x$
RHS $= 2 (c o s (3 x) + c o s x) - c o s 3 x$
RHS $= 2 c o s 3 x c o s 3 x + 2 c o s x c o s 3 x$
RHS $= 2 c o s^{2} 3 x + 2 c o s x c o s 3 x$
again
RHS $= 2 c o s^{2} 3 x + c o s 4 x + c o s 2 x$
RHS $= 2 (\frac{1 + c o s 2. (3 x)}{2}) + c o s 4 x + c o s 2 x$ { $c o s 2 A = 2 c o s^{2} A - 1, c o s^{2} A = \frac{1 + c o s 2 A}{2}$
RHS $= 1 + c o s 2 x + c o s 4 x + c o s 6 x =$ LHS
Hence proved

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General Solution of cos theta = cos alpha

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