Is sin^2(θ)= 3/4 sin(θ)-1/4 sin(3θ) ?

The answer to whether sin^2(θ)= 3/4 sin(θ)-1/4 sin(3θ) is False

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prove sin^2(θ)= 3/4 sin(θ)-1/4 sin(3θ)

prove

sin^{2} (θ) = \frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

False

Solution steps

sin^{2} (θ) = \frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

Show that the two sides are not equal

For

sin^{2} (θ) = \frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

plug in

θ = 1

sin^{2} (1) = 0.70807 \dots

sin^{2} (1)

Refine to a decimal form

= 0.70807 \dots

\frac{3}{4} sin (1) - \frac{1}{4} sin (3 \cdot 1) = 0.59582 \dots

\frac{3}{4} sin (1) - \frac{1}{4} sin (3 \cdot 1)

Refine to a decimal form

= 0.59582 \dots

The two sides are not equal

\Rightarrow False

\frac{1 + tan ^{2} ( x )}{cos ^{2} ( x )} = sec^{4} (x)

\frac{tan ^{2} ( x )}{sec ( x ) + 1} = sec (x)

sin^{3} (x) + cos^{3} (x) tan (x) = sin (x)

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

tan^{3} (x) = (tan (x))^{3}

Is sin^2(θ)= 3/4 sin(θ)-1/4 sin(3θ) ?
The answer to whether sin^2(θ)= 3/4 sin(θ)-1/4 sin(3θ) is False