Is tan(θ)sin(2θ)=2-2cos^2(θ) ?

The answer to whether tan(θ)sin(2θ)=2-2cos^2(θ) is True

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prove tan(θ)sin(2θ)=2-2cos^2(θ)

prove

tan (θ) sin (2 θ) = 2 - 2 cos^{2} (θ)

True

Solution steps

tan (θ) sin (2 θ) = 2 - 2 cos^{2} (θ)

Manipulating left side

tan (θ) sin (2 θ)

Express with sin, cos

sin (2 θ) tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

sin (2 θ) \frac{sin ( θ )}{cos ( θ )} : \frac{sin ( θ ) sin ( 2 θ )}{cos ( θ )}

sin (2 θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

\frac{2 sin ( θ ) cos ( θ ) sin ( θ )}{cos ( θ )} : 2 sin^{2} (θ)

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

2 sin (θ) cos (θ) sin (θ) = 2 sin^{2} (θ) cos (θ)

2 sin (θ) cos (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 2 cos (θ) sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 cos (θ) sin^{2} (θ)

= \frac{2 sin ^{2} ( θ ) cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= 2 sin^{2} (θ)

= 2 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 2 (1 - cos^{2} (θ))

= 2 (1 - cos^{2} (θ))

Expand

(1 - cos^{2} (θ)) \cdot 2 : 2 - 2 cos^{2} (θ)

(1 - cos^{2} (θ)) \cdot 2

= 2 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (θ)

= 2 \cdot 1 - 2 cos^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (θ)

= 2 - 2 cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

- 30 cos (3 x) sin (3 x) = - 15 sin (6 x)

2 sin^{2} (θ) - 1 = sin^{4} (θ) - cos^{4} (θ)

csc (2 θ) = \frac{1 + cot ^{2} ( θ )}{2 cot ( θ )}

csc^{4} (θ) - cot^{4} (θ) = 2 cot^{2} (θ) + 1

sec (b) cot (b) = csc (b)

Is tan(θ)sin(2θ)=2-2cos^2(θ) ?
The answer to whether tan(θ)sin(2θ)=2-2cos^2(θ) is True