Question

If `sin(2x) + sinx + cosx = 1`, then how do you show that `sinx +cosx = 1`?

Answer 1

Let's prove that the second expression equals the first.

If we square both sides, the equation becomes.

`(sinx + cosx)^2 = 1^2`

`sin^2x+ cos^2x + 2sinxcosx = 1`

Now use the identity `sin^2x + cos^2x = 1`.

`1 + 2sinxcosx = 1`

`2sinxcosx = 0`

We know also that `sin(2x) = 2sinxcosx`. If we substitute `sin(2x) = 0` into the first equation, and substitute the fact that `sinx + cosx = 1`, we get

`0 + 1 = 1`

Which is obviously true.

Hopefully this helps!

Answer 2

Please see below.

Explanation:

I will make use of `sin(2x) = 2sinxcosx`

Suppose that: `" "` `2sinxcosx+sinx+cosx = 1`

The we must have: `" "` `sinx+cosx=1-2sinxcosx`

Which implies that: `" "` `(sinx+cosx)^2=(1-2sinxcosx)^2`

So that: `" "` 1+2sinxcosx=1-4sinxcosx+4sin^2xcos^2x#.

Which assures us that: `" "` `6sinxcosx-4sin^2xcos^2x=0`

Factoring, we see that either

`sinxcosx=0` `" "` or else `" "` 6-4sinxcosx=0#

The second case requires that `sinxcosx=3/2` which is not possible. (`-1/2 <= sinx cosx <= 1/2` see note below.)

In the first case we have either

`sinx = 0` or `cosx = 0`

If `sinx = 0`, the we know `cosx = +-1` But the solution with `cosx = -1` is extraneous. (It does not solve the unsquared original equation.)

If `cosx = 0`, the we know `sinx = +-1` But the solution with `sinx = -1` is extraneous. (It does not solve the unsquared original equation.)

This leaves us with only two possibilities:

`sinx = 0` and `cosx = 1` OR `sinx =1` and `cosx=0`

In either case, the sum of `sinx` and `cosx` is `1`.

Note
`sinxcosx = 1/2sin(2x)`,

and `-1 <= sin(2x) <= 1`

so `-1/2 <= sinx cosx <= 1/2`