Is 1-tan^2(x/2)=(2cos(x))/(1+cos(x)) ?

The answer to whether 1-tan^2(x/2)=(2cos(x))/(1+cos(x)) is True

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prove 1-tan^2(x/2)=(2cos(x))/(1+cos(x))

Solution

prove

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

Solution

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating left side

1 - tan^{2} (u)

Express with sin, cos

1 - tan^{2} (u)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

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

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

Divide the numbers:

\frac{2}{2} = 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is 1-tan^2(x/2)=(2cos(x))/(1+cos(x)) ?
The answer to whether 1-tan^2(x/2)=(2cos(x))/(1+cos(x)) is True