Is (sin(x)}{sin(2x)}=\frac{sec(x))/2 ?

The answer to whether (sin(x)}{sin(2x)}=\frac{sec(x))/2 is True

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prove (sin(x)}{sin(2x)}=\frac{sec(x))/2

prove

\frac{sin ( x )}{sin ( 2 x )} = \frac{sec ( x )}{2}

True

Solution steps

\frac{sin ( x )}{sin ( 2 x )} = \frac{sec ( x )}{2}

Manipulating left side

\frac{sin ( x )}{sin ( 2 x )}

Rewrite using trig identities

\frac{sin ( x )}{sin ( 2 x )}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= \frac{sin ( x )}{2 sin ( x ) cos ( x )}

Cancel the common factor:

sin (x)

= \frac{1}{2 cos ( x )}

= \frac{1}{2 cos ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{1}{2 \cdot \frac{1}{s e c ( x )}}

Simplify

\frac{1}{2 \cdot \frac{1}{s e c ( x )}}

Multiply

2 \cdot \frac{1}{sec ( x )} : \frac{2}{sec ( x )}

2 \cdot \frac{1}{sec ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{sec ( x )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{sec ( x )}

= \frac{1}{\frac{2}{s e c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( x )}{2}

\frac{sec ( x )}{2}

\frac{sec ( x )}{2}

We showed that the two sides could take the same form

\Rightarrow True

sin (2 x) = 2 sin (x) \cdot cos (x)

tan (a) \frac{1}{sin ^{2} ( a )} = csc (a) sec (a)

(1 + csc (x)) (1 + csc (- x)) = - cot^{4} (x)

1 - \frac{sin ^{2} ( t )}{1 + cos ( t )} = cos (t)

(csc (α) + cot (α)) (sec (α) - 1) = tan (α)

Is (sin(x)}{sin(2x)}=\frac{sec(x))/2 ?
The answer to whether (sin(x)}{sin(2x)}=\frac{sec(x))/2 is True