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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

2 - sin (0) = 2

2 - 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

= 2 - 0

Simplify

= 2

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

4 - sin^{2} (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

= 4 - 0^{2}

Simplify

= 4

The two sides are not equal

\Rightarrow False

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\frac{cos ( - x )}{sin ( - x )} = - cot (x)

sin (\frac{9 π}{4}) = cos (\frac{9 π}{4})

tan (β) = csc (β) sec (β) - cot (β)

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