Is sin^4(t)-cos^4(t)=1-2cos^2(x) ?

The answer to whether sin^4(t)-cos^4(t)=1-2cos^2(x) is False

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prove sin^4(t)-cos^4(t)=1-2cos^2(x)

prove

sin^{4} (t) - cos^{4} (t) = 1 - 2 cos^{2} (x)

False

Solution steps

sin^{4} (t) - cos^{4} (t) = 1 - 2 cos^{2} (x)

Show that the two sides are not equal

For

sin^{4} (t) - cos^{4} (t) = 1 - 2 cos^{2} (x)

plug in

t = 0

For

sin^{4} (t) - cos^{4} (t) = 1 - 2 cos^{2} (x)

plug in

x = 1

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

sin^{4} (0) - cos^{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}

= 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^{4} - 1^{4}

Simplify

= - 1

1 - 2 cos^{2} (1) = 0.41614 \dots

1 - 2 cos^{2} (1)

Refine to a decimal form

= 0.41614 \dots

The two sides are not equal

\Rightarrow False

csc^{2} (A) = 1 + tan^{2} (A)

3 tan^{2} (x) = 3 (tan (x))^{2}

3 cos (θ) = 3 sin (- θ)

csc^{2} (θ) \cdot tan^{2} (θ) - 1 = tan^{2} (θ)

tan (a) + sec (a) = \frac{1 + sin ( a )}{cos ( a )}

Is sin^4(t)-cos^4(t)=1-2cos^2(x) ?
The answer to whether sin^4(t)-cos^4(t)=1-2cos^2(x) is False