Is tan^2(x)=2sin^2(x)-1 ?

The answer to whether tan^2(x)=2sin^2(x)-1 is False

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prove tan^2(x)=2sin^2(x)-1

prove

tan^{2} (x) = 2 sin^{2} (x) - 1

False

Solution steps

tan^{2} (x) = 2 sin^{2} (x) - 1

Show that the two sides are not equal

For

tan^{2} (x) = 2 sin^{2} (x) - 1

plug in

x = 0

tan^{2} (0) = 0

tan^{2} (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= 0^{2}

Simplify

= 0

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

2 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

= 2 \cdot 0^{2} - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

tan (- \frac{π}{2} - θ) = cot (θ)

cos (2 x) + tan (x) sin (2 x) = 1

\frac{( 1 + tan ( θ ) )}{( 1 + cot ( θ ) )} = tan (θ)

\frac{1}{sin ^{2} ( θ )} = 1 - tan^{2} (θ)

\frac{1}{( sec ( θ ) ) ^{2}} + \frac{1}{( csc ( θ ) ) ^{2}} = 1