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

sec (θ) - cos (θ) = tan (θ) sin (θ)

\frac{csc ( θ ) + sec ( θ )}{tan ( θ ) + 1} = csc (θ)

cot (x) cos (x) + sin (x) = csc (x)

sin (3 θ) = 3 sin (θ) - 4 sin^{3} (θ)

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

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