Is sin^2(θ/2)=(sec(θ)-1)/(2sec(θ)) ?

The answer to whether sin^2(θ/2)=(sec(θ)-1)/(2sec(θ)) is True

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prove sin^2(θ/2)=(sec(θ)-1)/(2sec(θ))

Solution

prove

sin^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{2 sec ( θ )}

Solution

True

Solution steps

sin^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{2 sec ( θ )}

Let:

u = \frac{θ}{2}

sin^{2} (u) = \frac{sec ( 2 u ) - 1}{2 sec ( 2 u )}

Prove

sin^{2} (u) = \frac{sec ( 2 u ) - 1}{2 sec ( 2 u )} :

True

sin^{2} (u) = \frac{sec ( 2 u ) - 1}{2 sec ( 2 u )}

Manipulating right side

\frac{sec ( 2 u ) - 1}{2 sec ( 2 u )}

Express with sin, cos

\frac{- 1 + sec ( 2 u )}{2 sec ( 2 u )}

Use the basic trigonometric identity:

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

= \frac{- 1 + \frac{1}{c o s ( 2 u )}}{2 \cdot \frac{1}{c o s ( 2 u )}}

Simplify

\frac{- 1 + \frac{1}{c o s ( 2 u )}}{2 \cdot \frac{1}{c o s ( 2 u )}} : \frac{- cos ( 2 u ) + 1}{2}

\frac{- 1 + \frac{1}{c o s ( 2 u )}}{2 \cdot \frac{1}{c o s ( 2 u )}}

Multiply

2 \cdot \frac{1}{cos ( 2 u )} : \frac{2}{cos ( 2 u )}

2 \cdot \frac{1}{cos ( 2 u )}

Multiply fractions:

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

= \frac{1 \cdot 2}{cos ( 2 u )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{cos ( 2 u )}

= \frac{- 1 + \frac{1}{c o s ( 2 u )}}{\frac{2}{c o s ( 2 u )}}

Join

- 1 + \frac{1}{cos ( 2 u )} : \frac{- cos ( 2 u ) + 1}{cos ( 2 u )}

- 1 + \frac{1}{cos ( 2 u )}

Convert element to fraction:

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

= - \frac{1 \cdot cos ( 2 u )}{cos ( 2 u )} + \frac{1}{cos ( 2 u )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ( 2 u ) + 1}{cos ( 2 u )}

Multiply:

1 \cdot cos (2 u) = cos (2 u)

= \frac{- cos ( 2 u ) + 1}{cos ( 2 u )}

= \frac{\frac{- c o s ( 2 u ) + 1}{c o s ( 2 u )}}{\frac{2}{c o s ( 2 u )}}

Divide fractions:

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

= \frac{( - cos ( 2 u ) + 1 ) cos ( 2 u )}{cos ( 2 u ) \cdot 2}

Cancel the common factor:

cos (2 u)

= \frac{- cos ( 2 u ) + 1}{2}

= \frac{1 - cos ( 2 u )}{2}

= \frac{1 - cos ( 2 u )}{2}

Rewrite using trig identities

\frac{1 - cos ( 2 u )}{2}

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

= \frac{1 - ( 1 - 2 sin ^{2} ( u ) )}{2}

Simplify

\frac{1 - ( 1 - 2 sin ^{2} ( u ) )}{2} : sin^{2} (u)

\frac{1 - ( 1 - 2 sin ^{2} ( u ) )}{2}

Expand

1 - (1 - 2 sin^{2} (u)) : 2 sin^{2} (u)

1 - (1 - 2 sin^{2} (u))

- (1 - 2 sin^{2} (u)) : - 1 + 2 sin^{2} (u)

- (1 - 2 sin^{2} (u))

Distribute parentheses

= - (1) - (- 2 sin^{2} (u))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (u)

= 1 - 1 + 2 sin^{2} (u)

1 - 1 = 0

= 2 sin^{2} (u)

= \frac{2 sin ^{2} ( u )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin^{2} (u)

= sin^{2} (u)

= sin^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

sin^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{2 sec ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sin^2(θ/2)=(sec(θ)-1)/(2sec(θ)) ?
The answer to whether sin^2(θ/2)=(sec(θ)-1)/(2sec(θ)) is True