Is 16((1-cos(2θ))/2)=8-8cos(2θ) ?

The answer to whether 16((1-cos(2θ))/2)=8-8cos(2θ) is True

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prove 16((1-cos(2θ))/2)=8-8cos(2θ)

prove

16 (\frac{1 - cos ( 2 θ )}{2}) = 8 - 8 cos (2 θ)

True

Solution steps

16 \cdot \frac{1 - cos ( 2 θ )}{2} = 8 - 8 cos (2 θ)

Manipulating left side

16 \frac{1 - cos ( 2 θ )}{2}

Expand

16 \cdot \frac{1 - cos ( 2 θ )}{2} : - 8 cos (2 θ) + 8

16 \cdot \frac{1 - cos ( 2 θ )}{2}

Multiply fractions:

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

= \frac{( 1 - cos ( 2 θ ) ) \cdot 16}{2}

Divide the numbers:

\frac{16}{2} = 8

= 8 (- cos (2 θ) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 8, b = - cos (2 θ), c = 1

= 8 (- cos (2 θ)) + 8 \cdot 1

Apply minus-plus rules

+ (- a) = - a

= - 8 cos (2 θ) + 8 \cdot 1

Multiply the numbers:

8 \cdot 1 = 8

= - 8 cos (2 θ) + 8

= 8 - 8 cos (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

cot (θ) sin (θ) cos (θ) = cos^{2} (θ)

sec (x) + tan (x) = \frac{sin ( x ) + 1}{cos ( x )}

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

tan (x) - 1 = cot (x)

sin^{2} (x) = \frac{sec ( x ) sin ( x )}{tan ( x ) + cot ( x )}

Is 16((1-cos(2θ))/2)=8-8cos(2θ) ?
The answer to whether 16((1-cos(2θ))/2)=8-8cos(2θ) is True