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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

cos (0) - cos^{3} (0)

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 - 1^{3}

Simplify

= 0

cos (0) - sin^{2} (0) = 1

cos (0) - sin^{2} (0)

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

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

= 1 - 0^{2}

Simplify

= 1

The two sides are not equal

\Rightarrow False

p ro v e cos (B) csc (B) tan (B) = 11

p ro v e tan (8 x) = \frac{8 tan ( x )}{1 - tan ^{2} ( x )}

p ro v e tan (- a) = \frac{cos ( \frac{π}{2} + a )}{cos ( a )}

p ro v e sin (θ) - \frac{1}{sin ( θ )} = cos (θ)

p ro v e cos^{8} (x) = sin^{7} (x) cos^{1} (x)

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