Is (sin^2(2x))/4 =sin^2(x)cos^2(x) ?

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

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prove (sin^2(2x))/4 =sin^2(x)cos^2(x)

prove

\frac{sin ^{2} ( 2 x )}{4} = sin^{2} (x) cos^{2} (x)

True

Solution steps

\frac{sin ^{2} ( 2 x )}{4} = sin^{2} (x) cos^{2} (x)

Manipulating left side

\frac{sin ^{2} ( 2 x )}{4}

Rewrite using trig identities

\frac{sin ^{2} ( 2 x )}{4}

Use the Double Angle identity:

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

= \frac{( 2 sin ( x ) cos ( x ) ) ^{2}}{4}

Simplify

\frac{( 2 sin ( x ) cos ( x ) ) ^{2}}{4} : sin^{2} (x) cos^{2} (x)

\frac{( 2 sin ( x ) cos ( x ) ) ^{2}}{4}

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

(2 sin (x) cos (x))^{2}

Apply exponent rule:

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

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

= \frac{2 ^{2} sin ^{2} ( x ) cos ^{2} ( x )}{4}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

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

Cancel the common factor:

2^{2}

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

20 cos^{2} (x) - cos (x) - 1 = 0

tan (\frac{y}{2}) = csc (y) - cot (y)

sin^{2} (θ) + 1 = - 2 - cos^{2} (x)

sec (x) \cdot cot (x) - sin (x) = cos (x) \cdot cot (x)

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

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