Is 4+4cos(2x)=8cos^2(x) ?

The answer to whether 4+4cos(2x)=8cos^2(x) is True

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prove 4+4cos(2x)=8cos^2(x)

prove

4 + 4 cos (2 x) = 8 cos^{2} (x)

True

Solution steps

4 + 4 cos (2 x) = 8 cos^{2} (x)

Manipulating left side

4 + 4 cos (2 x)

Rewrite using trig identities

4 + 4 cos (2 x)

Use the Double Angle identity:

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

= 4 + 4 (2 cos^{2} (x) - 1)

Simplify

4 + 4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x)

4 + 4 (2 cos^{2} (x) - 1)

Expand

4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 4

4 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

= 4 \cdot 2 cos^{2} (x) - 4 \cdot 1

Simplify

4 \cdot 2 cos^{2} (x) - 4 \cdot 1 : 8 cos^{2} (x) - 4

4 \cdot 2 cos^{2} (x) - 4 \cdot 1

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (x) - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 8 cos^{2} (x) - 4

= 8 cos^{2} (x) - 4

= 4 + 8 cos^{2} (x) - 4

Simplify

4 + 8 cos^{2} (x) - 4 : 8 cos^{2} (x)

4 + 8 cos^{2} (x) - 4

Group like terms

= 8 cos^{2} (x) + 4 - 4

4 - 4 = 0

= 8 cos^{2} (x)

= 8 cos^{2} (x)

= 8 cos^{2} (x)

= 8 cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( θ ) csc ( θ )}{2} = csc (2 θ)

\frac{cos ( x )}{cot ( x ) sin ( x )} = 1

\frac{1 + cos ( θ )}{csc ( θ )} = sin (θ) + cos (θ)

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

sin (2 θ) = 2 cos (θ) \cdot sin (θ)