Question

Cm đẳng thức sau : ....................................

\(\sin^6\left(\dfrac{x}{2}\right)-\cos^6\left(\dfrac{x}{2}\right)=\dfrac{1}{4}\cos x\left(\sin^2x-4\right)\)

Answer 1

Ta có:\(sin^6\left(\dfrac{x}{2}\right)-cos^6\left(\dfrac{x}{2}\right)\)=[sin²\(\left(\dfrac{x}{2}\right)\)-cos²\(\left(\dfrac{x}{2}\right)\)][sin⁴\(\left(\dfrac{x}{2}\right)\)+cos⁴\(\left(\dfrac{x}{2}\right)\)+sin²\(\left(\dfrac{x}{2}\right)\)+cos²\(\left(\dfrac{x}{2}\right)\)]=-cos(2.\(\dfrac{x}{2}\)){[sin²\(\left(\dfrac{x}{2}\right)\)+cos²\(\left(\dfrac{x}{2}\right)\)]²-sin²\(\left(\dfrac{x}{2}\right)\)cos²\(\left(\dfrac{x}{2}\right)\)}=-cosx[1-sin²\(\left(\dfrac{x}{2}\right)\)cos²\(\left(\dfrac{x}{2}\right)\)]

=-cosx[1-\(\dfrac{1}{4}\left(sin\left(\dfrac{x}{2}+\dfrac{x}{2}\right)+sin\left(\dfrac{x}{2}-\dfrac{x}{2}\right)\right)^2\)]

=-cosx(1-\(\dfrac{1}{4}sin^2x\))

=\(\dfrac{1}{4}cosx\left(sin^2x-4\right)\)