Is cos(x)+1(cos(x)-1)=cos^2(x)-1 ?

The answer to whether cos(x)+1(cos(x)-1)=cos^2(x)-1 is False

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prove cos(x)+1(cos(x)-1)=cos^2(x)-1

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

cos (0) + 1 \cdot (cos (0) - 1) = 1

cos (0) + 1 \cdot (cos (0) - 1)

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 \cdot (1 - 1)

Simplify

= 1

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

cos^{2} (0) - 1

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} - 1

Simplify

= 0

The two sides are not equal

\Rightarrow False

p ro v e sin (3 x) = 3 sin (x) - 4 sin (3 x)

p ro v e 2 cos^{2} (A) = 1 + cos (2 A)

p ro v e \frac{1 + cos ( a )}{sin ( a )} = cot (\frac{a}{s})

p ro v e cos (θ) = \frac{11}{6}

p ro v e 1 - sin^{2} (\frac{x}{4}) = cos (\frac{x}{2})

Is cos(x)+1(cos(x)-1)=cos^2(x)-1 ?
The answer to whether cos(x)+1(cos(x)-1)=cos^2(x)-1 is False