Is cos^2(θ/2)=(sec(θ)+1)/(2sec(θ)) ?

The answer to whether cos^2(θ/2)=(sec(θ)+1)/(2sec(θ)) is True

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prove cos^2(θ/2)=(sec(θ)+1)/(2sec(θ))

Solution

prove

cos^{2} (\frac{θ}{2}) = \frac{sec ( θ ) + 1}{2 sec ( θ )}

Solution

True

Solution steps

cos^{2} (\frac{θ}{2}) = \frac{sec ( θ ) + 1}{2 sec ( θ )}

Let:

u = \frac{θ}{2}

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

Prove

cos^{2} (u) = \frac{sec ( 2 u ) + 1}{2 sec ( 2 u )} :

True

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

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

Simplify

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

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

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

1 + 2 cos^{2} (u) - 1

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (u)

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

Divide the numbers:

\frac{2}{2} = 1

= cos^{2} (u)

= cos^{2} (u)

= cos^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

cos^{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 cos^2(θ/2)=(sec(θ)+1)/(2sec(θ)) ?
The answer to whether cos^2(θ/2)=(sec(θ)+1)/(2sec(θ)) is True