Is tan(2x)=tan(x+x) ?

The answer to whether tan(2x)=tan(x+x) is True

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prove tan(2x)=tan(x+x)

prove

tan (2 x) = tan (x + x)

True

Solution steps

tan (2 x) = tan (x + x)

Manipulating left side

tan (2 x)

Express with sin, cos

tan (2 x)

Use the basic trigonometric identity:

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

= \frac{sin ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x )}{cos ( 2 x )}

Rewrite using trig identities

\frac{sin ( 2 x )}{cos ( 2 x )}

Use the Double Angle identity:

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

= \frac{2 sin ( x ) cos ( x )}{cos ( 2 x )}

Use the Double Angle identity:

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

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

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

Manipulating right side

tan (x + x)

Rewrite using trig identities

tan (x + x)

Use the basic trigonometric identity:

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

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

Use the Angle Sum identity:

sin (s + s) = sin (s) cos (s) + cos (s) sin (s)

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

Use the Angle Sum identity:

cos (s + s) = cos (s) cos (s) - sin (s) sin (s)

= \frac{sin ( x ) cos ( x ) + cos ( x ) sin ( x )}{cos ( x ) cos ( x ) - sin ( x ) sin ( x )}

Simplify

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

\frac{sin ( x ) cos ( x ) + cos ( x ) sin ( x )}{cos ( x ) cos ( x ) - sin ( x ) sin ( x )}

Add similar elements:

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

= \frac{2 sin ( x ) cos ( x )}{cos ( x ) cos ( x ) - sin ( x ) sin ( x )}

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

cos (x) cos (x) - sin (x) sin (x)

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin^{3} (x) = \frac{3}{4} sin (x) - \frac{1}{4} sin (3 x)

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

cot (θ) (tan (θ) + cot (θ)) = csc^{2} (θ)

csc (θ) - cot (θ) cos (θ) = sin (θ)

2 cos (3 (\frac{11 π}{12})) = - 2