Is cos^2(θ)=sin^2(θ)-1 ?

The answer to whether cos^2(θ)=sin^2(θ)-1 is False

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prove cos^2(θ)=sin^2(θ)-1

prove

cos^{2} (θ) = sin^{2} (θ) - 1

False

Solution steps

cos^{2} (θ) = sin^{2} (θ) - 1

Show that the two sides are not equal

For

cos^{2} (θ) = sin^{2} (θ) - 1

plug in

θ = 0

cos^{2} (0) = 1

cos^{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

= 1^{2}

Simplify

= 1

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

sin^{2} (0) - 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

= 0^{2} - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

\frac{tan ( 2 θ ) + cot ( 2 θ )}{sec ( 2 θ )} = csc (2 θ)

\frac{sec ( θ )}{sec ( θ ) - 1} = \frac{1}{1 - cos ( θ )}

2 cos (θ) - \frac{cos ( 2 θ )}{cos ( θ )} = sec (θ)

cos (2 \cdot 30) = cos^{2} (30) - sin^{2} (30)

1 + tan^{2} (4 5^{\circ}) = sec^{2} (4 5^{\circ})