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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

sin (0) + cos (2 \cdot 0) = 1

sin (0) + cos (2 \cdot 0)

Simplify

= sin (0) + cos (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 + 1

Simplify

= 1

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

4 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

= 4 \cdot 0^{2} - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

\frac{sec ( x ) - cos ( x )}{sin ( x )} = tan (x)

tan (- θ) = tan (θ)

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

\frac{( 1 - cos ( a ) )}{1 + cos ( a )} = tan^{2} (\frac{a}{2})

cos (x - y) = cos (y - x)

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