Question

Write the following as product of trigonometric ratios?

(1) `cos3phi-cos4phi-cos5phi+cos6phi`
(2) `sin2alpha+sin4alpha+sin6alpha`
(3) `sin5phi-sin6phi-sin7phi+sin8phi`
(4) `sin4beta-2cos^2(2beta)+1`

Answer 1

(1) `4sin(phi/2)sinphicos((9phi)/2)`
(2) `4cosalphacos2alphasin3alpha`
(3) `4sin(phi/2)sinphisin((13phi)/2)`
(4) `sqrt2sin(4beta-pi/4)`

Explanation:

(1) `cos3phi-cos4phi-cos5phi+cos6phi`

= `(cos6phi-cos5phi)-(cos4phi-cos3phi)`

using `cosA-cosB=2sin((A+B)/2)sin((A-B)/2)` the above is equal to

`2sin((11phi)/2)sin(phi/2)-2sin((7phi)/2)sin(phi/2)`

= `2sin(phi/2){sin((11phi)/2)-sin((7phi)/2)}`

using `sinA-sinB=2cos((A+B)/2)sin((A-B)/2)` the above is equal to

= `2sin(phi/2){2cos((18phi)/4)sin((4phi)/4)}`

= `color(blue)(4sin(phi/2)sinphicos((9phi)/2)`

(2) `sin2alpha+sin4alpha+sin6alpha`

= `sin6alpha+sin2alpha+sin4alpha`

Using `sin2A=2sinAcosA` and `sinA+sinB=2sin((A+B)/2)cos((A-B)/2)`

= `2sin3alphacos3alpha+2sin3alphacosalpha`

= `2sin3alpha(cos3alpha+cosalpha)`

Using `cosA+cosB=2cos((A+B)/2)cos((A-B)/2)`, this is equal to

`2sin3alpha(2cos2alphacosalpha)`

= `color(blue)(4cosalphacos2alphasin3alpha)`

(3) `sin5phi-sin6phi-sin7phi+sin8phi` - use identities in (1) above

= `sin8phi-sin7phi-(sin6phi-sin5phi)`

= `2cos((15phi)/2)sin(phi/2)-2cos((11phi)/2)sin(phi/2)`

= `2sin(phi/2){cos((15phi)/2)-cos((11phi)/2)}`

= `2sin(phi/2){2sin((15phi+11phi)/4)sin((15phi-11phi)/4)}`

= `color(blue)(4sin(phi/2)sinphisin((13phi)/2))`

(4) `sin4beta-2cos^2(2beta)+1`

= `sin4beta-(2cos^2(2beta)-1)`

Using `sin2A=2sinAcosA` and `cos2A=2cos^2A-1`, this is equal to

`sin4beta-cos4beta` and as `sin(pi/2-A)=cosA` this is

= `sin4beta-sin(pi/2-4beta)` using identity in (1)

= `2cos((4beta+pi/2-4beta)/2)sin((4beta-pi/2+4beta)/2)`

= `2cos(pi/4)sin(4beta-pi/4)`

= `2/sqrt2sin(4beta-pi/4)`

= `color(blue)(sqrt2sin(4beta-pi/4))`