Is csc(2θ)=(csc(θ))/(2cos(θ)) ?

The answer to whether csc(2θ)=(csc(θ))/(2cos(θ)) is True

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prove csc(2θ)=(csc(θ))/(2cos(θ))

prove

csc (2 θ) = \frac{csc ( θ )}{2 cos ( θ )}

True

Solution steps

csc (2 θ) = \frac{csc ( θ )}{2 cos ( θ )}

Manipulating left side

csc (2 θ)

Express with sin, cos

csc (2 θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( 2 θ )}

= \frac{1}{sin ( 2 θ )}

Rewrite using trig identities

\frac{1}{sin ( 2 θ )}

Use the Double Angle identity:

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

= \frac{1}{2 sin ( θ ) cos ( θ )}

= \frac{1}{2 sin ( θ ) cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{2 \cdot \frac{1}{c s c ( θ )} cos ( θ )}

Simplify

\frac{1}{2 \cdot \frac{1}{c s c ( θ )} cos ( θ )}

Multiply

2 \cdot \frac{1}{csc ( θ )} cos (θ) : \frac{2 cos ( θ )}{csc ( θ )}

2 \cdot \frac{1}{csc ( θ )} cos (θ)

Multiply fractions:

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

= \frac{1 \cdot 2 cos ( θ )}{csc ( θ )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 cos ( θ )}{csc ( θ )}

= \frac{1}{\frac{2 c o s ( θ )}{c s c ( θ )}}

Apply the fraction rule:

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

= \frac{csc ( θ )}{2 cos ( θ )}

\frac{csc ( θ )}{2 cos ( θ )}

\frac{csc ( θ )}{2 cos ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

sin (- x) + csc (x) = cot (x) cos (x)

sin (x + y) = sin (x) cos (y) + cos (x) sin (y)

(sec (θ) + 1) (sec (θ) - 1) = tan^{2} (θ)

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

sin (x + y) = sin (x) cos (y) + sin (y) cos (x)

Is csc(2θ)=(csc(θ))/(2cos(θ)) ?
The answer to whether csc(2θ)=(csc(θ))/(2cos(θ)) is True