Solve equations using trigonometric identities

Products of sine and cosines of two angles can be expressed as sums, and sums can be expressed as products.

The starting point is the angle sum formula,
$\sin(A+B)=\sin A\cos B+\sin B\cos A\ \ \ \ \ \ \ \ \ (1)$sin(A+B)=sinAcosB+sinBcosA         (1).

From this we deduce

$\sin(A-B)$sin(A−B)	$=$=	$\sin A\cos(-B)+\sin(-B)\cos A$sinAcos(−B)+sin(−B)cosA
	$=$=	$\sin A\cos B-\sin B\cos A\ \ \ \ \ \ \ \ (2)$sinAcosB−sinBcosA        (2)

 

For the cosines of a sum and difference, we have

$\cos(A-B)$cos(A−B)	$=$=	$\sin\left(\frac{\pi}{2}-(A-B)\right)$sin(π2​−(A−B))
	$=$=	$\sin\left((\frac{\pi}{2}-A)+B\right)$sin((π2​−A)+B)
	$=$=	$\sin(\frac{\pi}{2}-A)\cos B+\sin B\cos(\frac{\pi}{2}-A)$sin(π2​−A)cosB+sinBcos(π2​−A)
	$=$=	$\cos A\cos B+\sin B\sin A\ \ \ \ \ (3)$cosAcosB+sinBsinA     (3)

And, from this we deduce

$\cos(A+B)$cos(A+B)	$=$=	$\cos\left(A-(-B)\right)$cos(A−(−B))
	$=$=	$\cos A\cos(-B)+\sin(-B)\sin A$cosAcos(−B)+sin(−B)sinA
	$=$=	$\cos A\cos B-\sin B\sin A\ \ \ \ \ \ \ \ (4)$cosAcosB−sinBsinA        (4)

 

Now, by adding equations $(1)$(1) and $(2)$(2), we get 
$\sin(A+B)+\sin(A-B)=2\sin A\cos B$sin(A+B)+sin(A−B)=2sinAcosB

That is, a product has been expressed as a sum:

$\sin A\cos B=\frac{1}{2}\left(\sin(A+B)+\sin(A-B)\right)\ \ \ \ \ \ \ \ \ \ (5)$sinAcosB=12​(sin(A+B)+sin(A−B))          (5)

Similarly, by adding equations $(3)$(3) and $(4)$(4) we get
$\cos(A-B)+\cos(A+B)=2\cos A\cos B$cos(A−B)+cos(A+B)=2cosAcosB.

That is:

$\cos A\cos B=\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right)\ \ \ \ \ \ \ \ \ \ (6)$cosAcosB=12​(cos(A−B)+cos(A+B))          (6).

By subtracting equation $(4)$(4) from equation $(3)$(3), we have
$\cos(A-B)-\cos(A+B)=2\sin B\sin A$cos(A−B)−cos(A+B)=2sinBsinA.

That is:

$\sin A\sin B=\frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)\ \ \ \ \ \ \ \ \ \ (7)$sinAsinB=12​(cos(A−B)−cos(A+B))          (7)

 

If these products can be expressed as sums, it must be possible to express sums as products. In each of the formulae $(5)$(5), $(6)$(6) and $(7)$(7), we can set $A+B=X$A+B=X and $A-B=Y$A−B=Y. Then, we must have $A=\frac{X+Y}{2}$A=X+Y2​ and $B=\frac{X-Y}{2}$B=X−Y2​.

After substituting for $A$A and $B$B in the three equations, we obtain:

$\sin X+\sin Y=2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)$sinX+sinY=2sinX+Y2​cosX−Y2​              (8)

$\cos X+\cos Y=2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)$cosX+cosY=2cosX+Y2​cosX−Y2​              (9)

$\cos Y-\cos X=2\sin\frac{X+Y}{2}\sin\frac{X-Y}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)$cosY−cosX=2sinX+Y2​sinX−Y2​              (10)

From $(8)$(8) we can argue that 

$\sin X-\sin Y=\sin X+\sin(-Y)=2\sin\frac{X-Y}{2}\cos\frac{X+Y}{2}\ \ \ \ \ \ \ \ \ (11)$sinX−sinY=sinX+sin(−Y)=2sinX−Y2​cosX+Y2​         (11)

 

Example 1

Verify that $\frac{\sin(7\theta)+\sin(5\theta)}{\cos(7\theta)-\cos(5\theta)}=-\cot\theta$sin(7θ)+sin(5θ)cos(7θ)−cos(5θ)​=−cotθ

Using formulae $(8)$(8) and $(10)$(10) we see that the left-hand side can be written

$\frac{2\sin\left(\frac{7\theta+5\theta}{2}\right)\cos\left(\frac{7\theta-5\theta}{2}\right)}{2\sin\left(\frac{7\theta+5\theta}{2}\right)\sin\left(\frac{5\theta-7\theta}{2}\right)}$2sin(7θ+5θ2​)cos(7θ−5θ2​)2sin(7θ+5θ2​)sin(5θ−7θ2​)​

$=\frac{\cos\theta}{-\sin\theta}$=cosθ−sinθ​

$=-\cot\theta$=−cotθ

 

Example 2

Solve $\cos3x+\cos2x=0$cos3x+cos2x=0.

Formula $(9)$(9) is applicable. The equation can be re-written as $2\cos\frac{3x+2x}{2}\cos\frac{3x-2x}{2}=0$2cos3x+2x2​cos3x−2x2​=0.

Thus, $2\cos\frac{5x}{2}\cos\frac{x}{2}=0$2cos5x2​cosx2​=0.

This has solutions where $\cos\frac{5x}{2}=0$cos5x2​=0 and where $\cos\frac{x}{2}=0$cosx2​=0.

So, $\frac{5x}{2}=\frac{\pi}{2}+2n\pi$5x2​=π2​+2nπ or $\frac{5x}{2}=\frac{3\pi}{2}+2n\pi$5x2​=3π2​+2nπ or $\frac{x}{2}=\frac{\pi}{2}+2n\pi$x2​=π2​+2nπ or $\frac{x}{2}=\frac{3\pi}{2}+2n\pi$x2​=3π2​+2nπ.

Therefore, the general solutions are
$x=\frac{4n+1}{5}\pi$x=4n+15​π
$x=\frac{4n+3}{5}\pi$x=4n+35​π
$x=(4n+1)\pi$x=(4n+1)π
$x=(4n+3)\pi$x=(4n+3)π

where $n$n is an integer.

In fact, we only need the first two solution expressions since the others are included in them. The graph of the function $y=\cos3x+\cos2x$y=cos3x+cos2x should help to visualise the locations of the zeros.

 

 

Worked Examples

Question 1

Question 2

Question 3