# why is sinx1 * sinx2 = -sinx1sinx2 in the equation (cosx1 *sinx1)(cosx2 * sinx2)if possible explain the equation fully

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### 1 Answer

If `(cos x_1*sin x_1)*(cos x_2*sin x_2)=0 =gt cos x_1*sin x_1 = 0` and `cos x_2*sin x_2 = 0`

If `cos x_1*sin x_1 = 0 =gt cos x_1= 0 ` or `sin x_1 = 0`

The sine and cosine functions cannot be cancelling out simultaneously, therefore if `cos x_1= 0, sin x_1 ` is `+-1 ` and if `sin x_1 = 0, cos x_1 ` is `+-1` .

`cos x_1= 0 =gt x_1 = pi/2 + kpi`

`sin x_1 = 0 =gt x_1 = kpi`

If `cos x_2*sin x_2 = 0 =gt cos x_2= 0 ` or `sin x_2 = 0`

`cos x_2= 0 =gt x_2 = pi/2 + kpi`

`sin x_2 = 0 =gt x_2 = kpi`

Let's check why `sin x_1*sin x_2=-sin x_1*sin x_2` , considering the conditions above.

`sin x_1*sin x_2=-sin x_1*sin x_2 lt=gt sin x_1*sin x_2+sin x_1*sin x_2 = 0`

`2 sin x_1*sin x_2 = 0 =gt sin x_1*sin x_2 = 0`

ANSWER: **Since the equations `sin x_1 = 0`**** and `sin x_2=0` have identical solutions (`x_1 ` = `x_2` ) =>`sin x_1*sin x_2=-sin x_1*sin x_2` **