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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

cos^{2} (0) - 4 = - 3

cos^{2} (0) - 4

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 1^{2} - 4

Simplify

= - 3

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

sec (x \cdot 0) = 1

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

\frac{2 cos ( x )}{csc ( x ) - 2 sin ( x )} = tan (2 x)

\frac{1 + csc ( a )}{cot ( a ) + cos ( a )} = sec (a)

sec (θ) - sec (θ) sin^{2} (θ) = cos (θ)