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

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

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

prove

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

True

Solution steps

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

Manipulating left 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 (θ))^{2} + (cos (θ))^{2} = 1

tan^{2} (π) = tan (π)

\frac{( - 1 + sec ^{2} ( b ) )}{tan ( b )} = tan (b)

cos (- θ + \frac{π}{2}) = sin (θ)

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

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