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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

\frac{1 - tan ^{2} ( B )}{1 + tan ^{2} ( B )}

Express with sin, cos

\frac{1 - tan ^{2} ( B )}{1 + tan ^{2} ( B )}

Use the basic trigonometric identity:

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

= \frac{1 - ( \frac{s i n ( B )}{c o s ( B )} ) ^{2}}{1 + ( \frac{s i n ( B )}{c o s ( B )} ) ^{2}}

Simplify

\frac{1 - ( \frac{s i n ( B )}{c o s ( B )} ) ^{2}}{1 + ( \frac{s i n ( B )}{c o s ( B )} ) ^{2}} : \frac{cos ^{2} ( B ) - sin ^{2} ( B )}{cos ^{2} ( B ) + sin ^{2} ( B )}

\frac{1 - ( \frac{s i n ( B )}{c o s ( B )} ) ^{2}}{1 + ( \frac{s i n ( B )}{c o s ( B )} ) ^{2}}

Apply exponent rule:

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

= \frac{1 - ( \frac{s i n ( B )}{c o s ( B )} ) ^{2}}{1 + \frac{s i n ^{2} ( B )}{c o s ^{2} ( B )}}

Apply exponent rule:

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

= \frac{1 - \frac{s i n ^{2} ( B )}{c o s ^{2} ( B )}}{1 + \frac{s i n ^{2} ( B )}{c o s ^{2} ( B )}}

Join

1 + \frac{sin ^{2} ( B )}{cos ^{2} ( B )} : \frac{cos ^{2} ( B ) + sin ^{2} ( B )}{cos ^{2} ( B )}

1 + \frac{sin ^{2} ( B )}{cos ^{2} ( B )}

Convert element to fraction:

1 = \frac{1 cos ^{2} ( B )}{cos ^{2} ( B )}

= \frac{1 \cdot cos ^{2} ( B )}{cos ^{2} ( B )} + \frac{sin ^{2} ( B )}{cos ^{2} ( B )}

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} ( B ) + sin ^{2} ( B )}{cos ^{2} ( B )}

Multiply:

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

= \frac{cos ^{2} ( B ) + sin ^{2} ( B )}{cos ^{2} ( B )}

= \frac{1 - \frac{s i n ^{2} ( B )}{c o s ^{2} ( B )}}{\frac{c o s ^{2} ( B ) + s i n ^{2} ( B )}{c o s ^{2} ( B )}}

Join

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

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

Convert element to fraction:

1 = \frac{1 cos ^{2} ( B )}{cos ^{2} ( B )}

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

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} ( B ) - sin ^{2} ( B )}{cos ^{2} ( B )}

Multiply:

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

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

= \frac{\frac{c o s ^{2} ( B ) - s i n ^{2} ( B )}{c o s ^{2} ( B )}}{\frac{c o s ^{2} ( B ) + s i n ^{2} ( B )}{c o s ^{2} ( B )}}

Divide fractions:

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

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

Cancel the common factor:

cos^{2} (B)

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( B ) - sin ^{2} ( B )}{cos ^{2} ( B ) + sin ^{2} ( B )}

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= cos^{2} (B) - sin^{2} (B)

Use the Double Angle identity:

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

= cos (2 B)

= cos (2 B)

We showed that the two sides could take the same form

\Rightarrow True

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