What is expand-(2cos(x)+1)(2cos(x)-1)sin^2(x) ?

The solution to expand-(2cos(x)+1)(2cos(x)-1)sin^2(x) is -4sin^2(x)cos^2(x)+sin^2(x)

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expand-(2cos(x)+1)(2cos(x)-1)sin^2(x)

expand

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

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

Solution steps

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

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

Apply Difference of Two Squares

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

- sin^{2} (x) \cdot 4 cos^{2} (x) - (- sin^{2} (x)) \cdot 1 : - 4 sin^{2} (x) cos^{2} (x) + sin^{2} (x)

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

\frac{1}{s \cdot ( s + 2 )}

(12 b^{20} - 9 m^{\frac{3}{2}})^{2}

\frac{s ^{2} + 1}{( s ^{2} + s + 10 ) s ^{2}}

t (x) x^{^{''}} + (5 t (x) - 5) x^{^{'}} + 5 x

\frac{54 ( x ^{2} + 3 )}{( x ^{2} - 9 ) ^{3}}

What is expand-(2cos(x)+1)(2cos(x)-1)sin^2(x) ?
The solution to expand-(2cos(x)+1)(2cos(x)-1)sin^2(x) is -4sin^2(x)cos^2(x)+sin^2(x)