91影视

Given identity is:

$1鈭�\cos 鈦�\left(2胃\right)+\cos 鈦�\left(4胃\right)鈭�\cos 鈦�\left(6胃\right)=4\sin 鈦�胃\cos 鈦�\left(2胃\right)\sin 鈦�\left(3胃\right)$

We have to establish the identity.

In proving the given identity, group the expression at the left hand side.

$1鈭�\cos 鈦�\left(2胃\right)+\cos 鈦�\left(4胃\right)鈭�\cos 鈦�\left(6胃\right)=\left[\cos \left(0\right)-\cos \left(2胃\right)\right]+\left[\cos \left(4胃\right)-\cos \left(6胃\right)\right]$

Simplify the groups in the LHS using the identity,

$\cos A-鈥�\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

Simplify the left hand side now,

$\begin{array}{r}1鈭�\cos 鈦�\left(2胃\right)+\cos 鈦�\left(4胃\right)鈭�\cos 鈦�\left(6胃\right)\\ =[\cos 鈦�0鈭�\cos 鈦�(6胃\left)\right]+[\cos 鈦�(4胃)+\cos 鈦�(2胃\left)\right]\\ =鈭�2\sin 鈦�\left(\frac{0+6胃}{2}\right)\sin 鈦�\left(\frac{0鈭�6胃}{2}\right)鈭�2\sin 鈦�\left(\frac{4胃+2胃}{2}\right)\sin 鈦�\left(\frac{4胃鈭�2胃}{2}\right)\\ =鈭�2\sin 鈦�\left(\frac{6胃}{2}\right)\sin 鈦�\left(\frac{鈭�6胃}{2}\right)鈭�2\sin 鈦�\left(\frac{6胃}{2}\right)\sin 鈦�\left(\frac{2胃}{2}\right)\\ =鈭�2\sin 鈦�\left(3胃\right)\sin 鈦�(鈭�3胃)鈭�2\sin 鈦�\left(3胃\right)\sin 鈦�\left(胃\right)\\ =鈭�2[\sin 鈦�(3胃)\sin 鈦�(鈭�3胃)+\sin 鈦�(3胃)\sin 鈦�(胃\left)\right]\\ =鈭�2\sin 鈦�\left(3胃\right)[\sin 鈦�(鈭�3胃)+\sin 鈦�(胃\left)\right]\\ =鈭�2\sin 鈦�\left(3胃\right)[\sin 鈦�(鈭�3胃)+\sin 鈦�(胃\left)\right]\end{array}$

Simplify the LHS further using the identity,

$\sin A-\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

Let role="math" localid="1646555577512" $$\begin{gathered} A=-3胃 \\ B=胃 \end{gathered}$$

role="math" localid="1646555684800" $\begin{array}{r}1鈭�\cos 鈦�\left(2胃\right)+\cos 鈦�\left(4胃\right)鈭�\cos 鈦�\left(6胃\right)=鈭�2\sin 鈦�\left(3胃\right)[\sin 鈦�(鈭�3胃)+\sin 鈦�(胃\left)\right]\\ =鈭�2\sin 鈦�\left(3胃\right)\left[2\sin 鈦�\left(\frac{鈭�3胃+胃}{2}\right)\cos 鈦�\left(\frac{鈭�3胃鈭�胃}{2}\right)\right]\\ =鈭�2\sin 鈦�\left(3胃\right)\left[2\sin 鈦�\left(\frac{鈭�2胃}{2}\right)\cos 鈦�\left(\frac{鈭�4胃}{2}\right)\right]\\ =鈭�2\sin 鈦�\left(3胃\right)[2\sin 鈦�(鈭�胃)\cos 鈦�(鈭�2胃\left)\right]\\ =鈭�2\sin 鈦�\left(3胃\right)[鈭�2\sin 鈦�胃\cos 鈦�(2胃\left)\right]\\ =4\sin 鈦�胃\cos 鈦�\left(2胃\right)\sin 鈦�\left(3胃\right)\end{array}$