Is sin^2(2θ)=4sin^2(θ)cos^2(θ) ?

The answer to whether sin^2(2θ)=4sin^2(θ)cos^2(θ) is True

prove sin^2(2θ)=4sin^2(θ)cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (2 θ)

Rewrite using trig identities

sin^{2} (2 θ)

Use the Double Angle identity:

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

= (2 sin (θ) cos (θ))^{2}

Simplify

(2 sin (θ) cos (θ))^{2} : 4 sin^{2} (θ) cos^{2} (θ)

(2 sin (θ) cos (θ))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} sin^{2} (θ) cos^{2} (θ)

2^{2} = 4

= 4 sin^{2} (θ) cos^{2} (θ)

= 4 sin^{2} (θ) cos^{2} (θ)

= 4 sin^{2} (θ) cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{cos ( 4 x ) + cos ( 6 x )}{sin ( 4 x ) - sin ( 6 x )} = - cot (x)

p ro v e 2 sin^{2} (x) = (1 - cos (2 x))

p ro v e 2 sin (4 x) = 4 sin (2 x) cos (2 x)

p ro v e \frac{1}{cos ^{2} ( x )} = (\frac{1}{cos ( x )})^{2}

p ro v e 1 - \frac{( cos ^{2} ( θ ) )}{1 + sin ( θ )} = sin (θ)

Is sin^2(2θ)=4sin^2(θ)cos^2(θ) ?
The answer to whether sin^2(2θ)=4sin^2(θ)cos^2(θ) is True