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

The answer to whether prove-sin^2(x)+cos^2(x)=cos(2x) is True

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

prove

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

True

Solution steps

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

Manipulating right side

cos (2 x)

Rewrite using trig identities

cos (2 x)

Rewrite as

= cos (x + x)

Use the Angle Sum identity:

cos (s + s) = cos (s) cos (s) - sin (s) sin (s)

= cos (x) cos (x) - sin (x) sin (x)

= cos (x) cos (x) - sin (x) sin (x)

Simplify

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

cos (x) cos (x) - sin (x) sin (x)

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin (x) + cos (x) = \frac{1}{cot ( x )}

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

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

tan (\frac{41 π}{2} + x) + cot (π + x) = 0

cot (\frac{θ}{2}) = \frac{sin ( θ )}{1 - cos ( θ )}

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