Prove that 1+sin(x)-cos(x) = 2sin(x/2)cos(x/2)+sin(x/2)

Question

I have not managed to establish how sin(x/2) relates to cos(x)

Answer 1

Trigonometric Identities we should know to begin:

 

(1) sin(x +y) = sin(x) cos(y) + cos(x) sin(y)

(2) 1 - cos(x) = 2 sin²(x/2)

 

Let's manipulate identity (1) -

sin(x + x) = sin(2x) = sin(x)cos(x) + cos(x)sin(x) = 2sin(x)cos(x)

 

So, sin(2x) = 2sin(x)cos(x)

 

We can divide what's in the parenthesis by 2.

 

->  sin(2x/2) = 2sin(x/2)cos(x/2)

->  sin(x) = 2sin(x/2)cos(x/2)

 

Now, let's look at the equation to be proven -

1 + sin(x) - cos(x) = sin(x) + 1 - cos(x)

 

Let's substitute our trig identities using what we got with trig identity (1) and (2) -

 

sin(x) + 1 - cos(x) = 2sin(x/2)cos(x/2) + 2sin²(x/2)

                          

                            = 2 sin(x/2) [cos(x/2) + sin(x/2)]

 

I hope this helps.

 

 

 

 

Answer 2

 

1-cos(x) = 2sin^2(x/2)

sin(x) = 2sin(x/2)cos(x/2)

So,

1+sin(x)-cos(x) = sin(x) + 1-cos(x)

=> 1+sin(x)-cos(x) = 2sin(x/2)cos(x/2) + 2sin^2(x/2)

=> 1+sin(x)-cos(x) = 2sin(x/2)(cos(x/2)+sin(x/2))

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Attn: This is an identity. Also, you need to add a parenthesis.