Is cos^8(x)=sin^7(x)cos^1(x) ?

The answer to whether cos^8(x)=sin^7(x)cos^1(x) is False

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prove cos^8(x)=sin^7(x)cos^1(x)

prove

cos^{8} (x) = sin^{7} (x) cos^{1} (x)

False

Solution steps

cos^{8} (x) = sin^{7} (x) cos^{1} (x)

Show that the two sides are not equal

For

cos^{8} (x) = sin^{7} (x) cos^{1} (x)

plug in

x = 0

cos^{8} (0) = 1

cos^{8} (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^{8}

Simplify

= 1

sin^{7} (0) cos^{1} (0) = 0

sin^{7} (0) cos^{1} (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

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

= 0^{7} \cdot 1^{1}

Simplify

= 0

The two sides are not equal

\Rightarrow False

p ro v e (cot (x) + 1)^{2} - 2 cot (x) = csc^{2} (x)

p ro v e cos (θ) = \frac{1}{3}

p ro v e sec (θ) - \frac{tan ( θ )}{csc ( θ )} = cos (θ)

p ro v e sec^{2} (x) = (1 + tan^{2} (x))

p ro v e cot^{2} (x) sec^{2} (x) = cot^{2} (x) + 1

Is cos^8(x)=sin^7(x)cos^1(x) ?
The answer to whether cos^8(x)=sin^7(x)cos^1(x) is False