Question

prove sin^2Q+cos^Q=1​

Answer 1

Answer:

1. sin²q+cos²q
2. since we know that; sin²q=1-cos²q on putting this vale in the given equation
3. 1-cos²q+cos²q
4. =1
5. hence proved
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Answer 2

We can define(!) the (first only R→R) functions sin and cos via exp(it)=cost+isint and the (complex) exponential as unique(!) solution of the differential equation f′(z)=f(z) with f(0)=1. We need only a few properties of exp that quickly follow from uniqueness of the solution:

Since z↦1expaexp(z+a) is also a solution whenever exp(a)≠0, we conclude by uniqueness that exp(a+b)=exp(a)exp(b) whenever exp(a)≠0.

Specifically, exp(a)=0 implies exp(a/2)=0, hence exp(2−na)=0. As exp(0)≠0 and 2−na→0 and exp is continous, we conclude exp(a)≠0 for all a. Therefore exp(a+b)=exp(a)exp(b) for all a,b.

Since z↦exp(z¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is also a solution, we conclude expz¯¯¯=expz¯¯¯¯¯¯¯¯¯¯¯ for all z.

This makes

cos2t+sin2t=(cost+isint)(cost−isint)=exp(it)⋅exp(it)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=exp(it)⋅exp(it¯¯¯¯)=exp(it)⋅exp(−it)=exp(it−it)=exp(0)=1.

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