Rewriting linear combinations of sin and cos

Here is a graph of the curves $y=\sin x$y=sinx and $y=\cos x$y=cosx on the same set of axes:

We have looked quite thoroughly at these functions and their graphs. For instance, we know that they are periodic, each with a period of $2\pi$2π, and have an amplitude of $1$1 centred around a height of $y=0$y=0. We are now going to look into sums and differences of $\sin x$sinx and $\cos x$cosx, using our knowledge of certain trigonometric identities.

 

$y=\sin x+\cos x$y=sinx+cosx

Let's begin by looking at the function $y=\sin x+\cos x$y=sinx+cosx. Using technology, we can obtain a graph of this function:

Notice that this looks much the same as a single sin or cos function, though it has a different amplitude and has been shifted horizontally. In fact we can rewrite the sum $\sin x+\cos x$sinx+cosx as a single trigonometric function. To see this we'll start with the variable $x$x and another fixed angle, which we choose to call $\alpha$α, and use the identity:

$\sin\left(x+\alpha\right)=\cos\alpha\sin x+\sin\alpha\cos x$sin(x+α)=cosαsinx+sinαcosx.

Now if $\cos\alpha$cosα and $\sin\alpha$sinα were to have the same value, we would be able to divide both sides by that value to leave the sum $\sin x+\cos x$sinx+cosx on the right hand side.

If we look back to the graph of the curves $y=\sin x$y=sinx and $y=\cos x$y=cosx at the beginning, we can see that the two graphs intersect at the angle $\frac{\pi}{4}$π4​. That is,

$\sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$sin(π4​)=cos(π4​)=1√2​.

So replacing $\alpha$α with $\frac{\pi}{4}$π4​, we have that

$\sin\left(x+\frac{\pi}{4}\right)$sin(x+π4​)	$=$=	$\cos\frac{\pi}{4}\sin x+\sin\frac{\pi}{4}\cos x$cosπ4​sinx+sinπ4​cosx
	$=$=	$\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x$1√2​sinx+1√2​cosx

 

Multiplying both sides by $\sqrt{2}$√2 gives us the result:

$\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$sinx+cosx=√2sin(x+π4​)

Here is a graph of $y=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$y=√2sin(x+π4​):

Comparing this to the graph of $y=\sin x+\cos x$y=sinx+cosx above, we can see that the two graphs (and therefore the two functions) are indeed the same!

 

$y=a\sin x+b\cos x$y=asinx+bcosx

We can generalise this process to any linear combination of $\sin x$sinx and $\cos x$cosx. That is, we can rewrite any function of the form $y=a\sin x+b\cos x$y=asinx+bcosx in the form $y=R\sin\left(x+\alpha\right)$y=Rsin(x+α), where $a$a, $b$b, $R$R and $\alpha$α are constants.

Using the same trigonometric identity, we have that $R\sin\left(x+\alpha\right)=R\cos\alpha\sin x+R\sin\alpha\cos x$Rsin(x+α)=Rcosαsinx+Rsinαcosx. Notice that this is in the form $a\sin x+b\cos x$asinx+bcosx, where

$a$a	$=$=	$R\cos\alpha$Rcosα	, and
$b$b	$=$=	$R\sin\alpha$Rsinα

 

We can rearrange these two equations to obtain expressions for $R$R and $\alpha$α in terms of $a$a and $b$b.

To find $R$R, we can square and add the two equations:

$a^2+b^2$a2+b2	$=$=	$R^2\cos^2\left(\alpha\right)+R^2\sin^2\left(\alpha\right)$R2cos2(α)+R2sin2(α)
	$=$=	$R^2\left(\sin^2\left(\alpha\right)+\cos^2\left(\alpha\right)\right)$R2(sin2(α)+cos2(α))
	$=$=	$R^2$R2

 

So we have that

$R=\sqrt{a^2+b^2}$R=√a2+b2.

To find $\alpha$α, we can divide the two equations:

$\frac{b}{a}$ba​	$=$=	$\frac{R\sin\alpha}{R\cos\alpha}$RsinαRcosα​
	$=$=	$\frac{\sin\alpha}{\cos\alpha}$sinαcosα​
	$=$=	$\tan\alpha$tanα

 

We can then find the value of $\alpha$α from this relation, though we may need to consider which quadrant the angle is in depending on the signs of $a$a and $b$b, since after all we know that $a=R\cos\alpha$a=Rcosα and $b=R\sin\alpha$b=Rsinα.

For example: if $\frac{b}{a}$ba​ was positive but the values of $a$a and $b$b were negative, then the value of $\alpha$α will lie in the third quadrant (since this is where $\tan$tan is positive, but sine and cosine are negative).

We can also rewrite a function of the form $y=a\sin x+b\cos x$y=asinx+bcosx in the form $y=R\cos\left(x+\alpha\right)$y=Rcos(x+α). Performing some similar rearrangements, we find in this case that $R=\sqrt{a^2+b^2}$R=√a2+b2 and $\tan\alpha=\frac{-a}{b}$tanα=−ab​.

 

 

Practice questions

Question 1

Question 2

Question 3