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+Y2cosX−Y2 (8)
$\cos X+\cos Y=2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)$cosX+cosY=2cosX+Y2cosX−Y2 (9)
$\cos Y-\cos X=2\sin\frac{X+Y}{2}\sin\frac{X-Y}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)$cosY−cosX=2sinX+Y2sinX−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−Y2cosX+Y2 (11)
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
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+2x2cos3x−2x2=0.
Thus, $2\cos\frac{5x}{2}\cos\frac{x}{2}=0$2cos5x2cosx2=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+15π
$x=\frac{4n+3}{5}\pi$x=4n+35π
$x=(4n+1)\pi$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.
Use the product-to-sum identities to express $2\cos\left(3x\right)\cos\left(9x\right)$2cos(3x)cos(9x) as a sum or difference of two trigonometric functions.
Express $\sin\left(6x\right)+\sin\left(4x\right)$sin(6x)+sin(4x) as a product of two trigonometric functions.
Evaluate $\cos195^\circ-\cos75^\circ$cos195°−cos75° by first expressing it as a product of two trigonometric functions.