# prove anglesangle 1+angle 2=90 degree sin 2 angle 1+ sin 2 angle 2 = 2 cos (angle 1+angle 2)

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You should come up with the substitution `angle 1 = alpha ` and `angle 2 = beta` such that:

`sin 2alpha + sin 2beta = 2 cos(alpha + beta)`

You should convert the sum `sin 2alpha + sin 2beta ` in a product such that:

`sin 2alpha + sin 2beta = 2 sin ((2alpha + 2beta)/2)*(cos (2alpha- 2beta)/2)`

Factoring out 2 to the right side yields:

`sin 2alpha + sin 2beta = 2 sin (2(alpha + beta)/2)*(cos 2(alpha- beta)/2)`

Reducing by 2 yields:

`sin 2alpha + sin 2beta = 2 sin (alpha + beta)*cos (alpha - beta)`

You need to use the information the problem provides: `alpha + beta = 90^o` , hence, you need to substitute `90^o` for `alpha + beta ` in product such that:

`sin 2alpha + sin 2beta = 2 sin 90^o*cos (alpha - beta)`

You should substitute 1 for `sin 90^o` such that:

`sin 2alpha + sin 2beta = 2 *cos (alpha - beta)`

**Hence, the last line proves the identity `sin 2alpha + sin 2beta = 2 *cos (alpha - beta)` , under the given condition `alpha + beta = 90^o` . The identity as stated in the original question is incorrect.**