Is sin(3x)=sin(3-4sin^2(x)) ?

The answer to whether sin(3x)=sin(3-4sin^2(x)) is False

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prove sin(3x)=sin(3-4sin^2(x))

prove

sin (3 x) = sin (3 - 4 sin^{2} (x))

False

Solution steps

sin (3 x) = sin (3 - 4 sin^{2} (x))

Show that the two sides are not equal

For

sin (3 x) = sin (3 - 4 sin^{2} (x))

plug in

x = 0

sin (3 \cdot 0) = 0

sin (3 \cdot 0)

Simplify

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0

sin (3 - 4 sin^{2} (0)) = 0.14112 \dots

sin (3 - 4 sin^{2} (0))

Refine to a decimal form

= 0.14112 \dots

The two sides are not equal

\Rightarrow False

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

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

cot (x) - 1 = \frac{csc ( x ) - sec ( x )}{sec ( x )}

tan (θ) = 2

sec (x) = sec (- x)

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