Is cos(2θ)=cos^{(2)}(θ)-sin^{(2)}(θ) ?

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

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

prove

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

True

Solution steps

cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

Manipulating left side

cos (2 θ)

Rewrite using trig identities

cos (2 θ)

Rewrite as

= cos (θ + θ)

Use the Angle Sum identity:

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

= cos (θ) cos (θ) - sin (θ) sin (θ)

= cos (θ) cos (θ) - sin (θ) sin (θ)

Simplify

cos (θ) cos (θ) - sin (θ) sin (θ) : cos^{2} (θ) - sin^{2} (θ)

cos (θ) cos (θ) - sin (θ) sin (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

2 sec (β) = 2 \cdot \frac{1}{cos ( β )}

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

csc^{2} (x) sec^{2} (x) = sec^{2} (x) + csc^{2} (x)

cot (s) = \frac{( sin ( 2 s ) )}{( 1 - cos ( 2 s ) )}

\frac{3 sin ^{2} ( x )}{tan ^{2} ( x )} = 3 cos^{2} (x)

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