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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

cos^{4} (sin^{4} (0)) = 1

cos^{4} (sin^{4} (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}

= cos^{4} (0^{4})

= cos^{4} (0^{4})

Simplify

= cos^{4} (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^{4}

Simplify

= 1

The two sides are not equal

\Rightarrow False

sec^{2} (x) + tan^{2} (x) = 1 + 2 tan^{2} (x)

\frac{sin ( a + b )}{sin ( a - b )} = \frac{1 + cot ( a ) tan ( b )}{1 - cot ( a ) tan ( b )}

cot (x + π) = cot (x)

(cos (x) + tan (x))^{2} = csc^{2} (x) sec^{2} (x)

tan (x) (2 - cos (x))^{2} = 0