Addition of sine and cosine

Question

 

 

 

 

 

      there doesn't have to be a bibliography. please the mark allocation for the info needed

Answer 1

ADDITION OF SINE AND COSINE

(1) We can use trig identities to bring the two components together. First let’s generalise then work towards specifics.

sin(a+b)=sin(a)cos(b)+cos(a)sin(b),

sin(a-b)=sin(a)cos(b)-cos(a)sin(b).

If we add these identities we get:

sin(a+b)+sin(a-b)=2sin(a)cos(b).

We can also use a relationship between sine and cosine:

sin(a-b)=cos(90-(a-b))=cos(90-a+b).

Therefore, sin(a+b)+cos(90-a+b)=2sin(a)cos(b).

Let x=a+b and y=90-a+b.

So x+y=90+2b, making b=½(x+y)-45.

From this a=x-b=½(x-y)+45.

So sin(x)+cos(y)=2sin(½(x-y)+45)cos(½(x+y)-45).

That’s the general case. Now to find sin(x)+cos(x) by plugging in y=x:

sin(x)+cos(x)=2sin(45)cos(x-45)=√2cos(x-45).

Now we use the fact that cos(θ)=cos(-θ) (cosine is an even function).

So sin(x)+cos(x)=√2cos(45-x).

cos(45-x)=sin(90-(45-x))=sin(x+45), and:

sin(x)+cos(x)=√2sin(x+45).

Comments

This problem concerns trigonometric identities. The understanding of these is fundamental to the solution.

Another shorter way to prove h(x)=f(x)+g(x)=sin(x)+cos(x)=√2sin(x+45):

sin(x+45)=sin(x)cos(45)+cos(x)sin(45).

sin(45)=cos(45)=√2/2, so:

sin(x+45)=sin(x)√2/2+cos(x)√2/2=√2/2(sin(x)+cos(x)).

Therefore, h(x)=√2sin(x+45)=√2(√2/2)(sin(x)+cos(x))=sin(x)+cos(x).