Is sin(x)(1+cos(2x))=sin(2x)cos(x) ?

The answer to whether sin(x)(1+cos(2x))=sin(2x)cos(x) is True

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

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

Simplify

1 + 2 cos^{2} (x) - 1 : 2 cos^{2} (x)

1 + 2 cos^{2} (x) - 1

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (x)

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

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

Manipulating right side

sin (2 x) cos (x)

Rewrite using trig identities

sin (2 x) cos (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (x) cos (x) cos (x)

Simplify

2 sin (x) cos (x) cos (x) : 2 cos^{2} (x) sin (x)

2 sin (x) cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

(sec (x) - tan (x)) (1 + sin (x)) = cos (x)

tan (- x) = cos (x)

\frac{1}{cos ( x )} - cos (x) = tan (x) sin (x)

cos (4 5^{\circ} - x) = cos (x - 4 5^{\circ})

\frac{1 - csc ^{2} ( A )}{csc ^{2} ( A )} = - cos^{2} (A)

Is sin(x)(1+cos(2x))=sin(2x)cos(x) ?
The answer to whether sin(x)(1+cos(2x))=sin(2x)cos(x) is True