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Range of Trigonometric Ratios from 0 to 90 Degrees

cos 4x. cos 8...

Question

$cos 4 x . cos 8 x - cos 5 x . cos 9 x = 0$ , if

A

c o s 12 x = c o s 14 x

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B

s i n 13 x = 0

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C

s i n x = 0

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D

c o s x = 0

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Solution

The correct options are
A $c o s 12 x = c o s 14 x$
B $s i n 13 x = 0$
C $s i n x = 0$
$c o s 4 x . c o s 8 x - c o s 5 x . c o s 9 x = 0 \Rightarrow c o s 4 x . c o s 8 x = c o s 5 x . c o s 9 x \Rightarrow 2 c o s 4 x . c o s 8 x = 2 c o s 5 x . c o s 9 x$
Using $2 c o s A c o s B = c o s (A - B) + c o s (A + B)$ , we get
$c o s (- 4 x) + c o s 12 x = c o s (- 4 x) + c o s 14 x \Rightarrow c o s 12 x = c o s 14 x \Rightarrow c o s 12 x - c o s 14 x = 0$
Hence option A is true
Now using $c o s A - c o s B = - 2 s i n (\frac{A + B}{2}) s i n (\frac{A - B}{2})$ , we get
$- 2 s i n 13 x s i n (- x) = 0 \Rightarrow 2 s i n 13 x s i n x = 0 \Rightarrow s i n 13 x = 0 s i n x = 0$
Hence option B and C are true

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