Is sin(2α)=(2tan(α))/(1+tan^2(α)) ?

The answer to whether sin(2α)=(2tan(α))/(1+tan^2(α)) is True

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Popular Trigonometry >

prove sin(2α)=(2tan(α))/(1+tan^2(α))

Solution

prove

sin (2 α) = \frac{2 tan ( α )}{1 + tan ^{2} ( α )}

Solution

True

Solution steps

sin (2 α) = \frac{2 tan ( α )}{1 + tan ^{2} ( α )}

Manipulating right side

\frac{2 tan ( α )}{1 + tan ^{2} ( α )}

Express with sin, cos

\frac{2 tan ( α )}{1 + tan ^{2} ( α )}

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{2 \cdot \frac{s i n ( α )}{c o s ( α )}}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

Apply exponent rule:

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

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

Multiply

2 \cdot \frac{sin ( α )}{cos ( α )} : \frac{2 sin ( α )}{cos ( α )}

2 \cdot \frac{sin ( α )}{cos ( α )}

Multiply fractions:

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

= \frac{sin ( α ) \cdot 2}{cos ( α )}

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

Apply the fraction rule:

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

= \frac{sin ( α ) \cdot 2}{cos ( α ) ( 1 + \frac{s i n ^{2} ( α )}{c o s ^{2} ( α )} )}

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot cos^{2} (α) = cos^{2} (α)

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

= \frac{2 sin ( α )}{\frac{c o s ^{2} ( α ) + s i n ^{2} ( α )}{c o s ^{2} ( α )} cos ( α )}

Multiply

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

cos (α) \frac{cos ^{2} ( α ) + sin ^{2} ( α )}{cos ^{2} ( α )}

Multiply fractions:

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

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

Cancel the common factor:

cos (α)

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

= \frac{2 sin ( α )}{\frac{c o s ^{2} ( α ) + s i n ^{2} ( α )}{c o s ( α )}}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

\frac{2 cos ( α ) sin ( α )}{cos ^{2} ( α ) + sin ^{2} ( α )}

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

= \frac{sin ( 2 α )}{1}

Apply rule

\frac{a}{1} = a

= sin (2 α)

= sin (2 α)

We showed that the two sides could take the same form

\Rightarrow True

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Is sin(2α)=(2tan(α))/(1+tan^2(α)) ?
The answer to whether sin(2α)=(2tan(α))/(1+tan^2(α)) is True