Question

Cos45°-cos75°\sin45°+sin75°=?

Answer 1

`2-sqrt3`

Explanation:

`(Cos45°-cos75°)/(sin45°+sin75°)`

Can be rewritten as:
`(Cos45°-cos(30°+45°))/(sin45°+sin(30°+45°)`

Applying cosine and sine sum identities:
`(Cos45°-(cos30°cos45°-sin30°sin45°))/(sin45°+(sin30°cos45°+cos30°sin45°))`

Simplify
`(sqrt2/2-(sqrt3/2*sqrt2/2-1/2*sqrt2/2))/(sqrt2/2+(1/2*sqrt2/2+sqrt3/2*sqrt2/2))=`

`(sqrt2/2-((sqrt6-sqrt2)/4))/(sqrt2/2+((sqrt2+sqrt6)/4))=`

`((2sqrt2)/4+(-sqrt6+sqrt2)/4)/((2sqrt2)/4+(sqrt2+sqrt6)/4)=`

`((3sqrt2-sqrt6)/4)/((3sqrt2+sqrt6)/4)=`

`(3sqrt2-sqrt6)/cancel4*cancel4/(3sqrt2+sqrt6)=`

`(3sqrt2-sqrt6)/(3sqrt2+sqrt6)`

We can rationalize:
`(3sqrt2-sqrt6)/(3sqrt2+sqrt6)*(3sqrt2-sqrt6)/(3sqrt2-sqrt6)=`

`(18-12sqrt3+6)/(18-6)=`

`(24-12sqrt3)/12=`

`2-sqrt3`

Answer 2

`2-sqrt3`

Explanation:

`"to obtain the exact value for the expression use the"`
`color(blue)"trigonometric identities"`

`•color(white)(x)cosC-cosD=-2sin((C+D)/2)sin((C-D)/2)`

`•color(white)(x)sinC+sinD=2sin((C+D)/2)cos((C-D)/2)`

`•color(white)(x)sin(x+-y)=sinxcosy+-cosxsiny`

`•color(white)(x)cos(x+-y)=cosxcosy∓sinxsiny`

`color(blue)"simplifying the numerator"`

`"with "C=45" and "D=75`

`cos45-cos75=-2sin60sin(-15)=2sin60sin15`

`"to obtain the exact value for "sin15`

`sin15=sin(45-30)=sin45cos30-cos45sin30`

`color(white)(sin15)=1/sqrt2xxsqrt3/2-1/sqrt2xx1/2`

`color(white)(sin15)=(sqrt3-1)/(2sqrt2)=1/4(sqrt6-sqrt2)`

`rArrcos45-cos75=2xxsqrt3/2xx1/4(sqrt6-sqrt2)`

`color(white)(xxxxxxxxxxxx)=1/4sqrt3(sqrt6-sqrt2)`

`color(blue)"simplifying the denominator"`

`"with "C=45" and "D=75`

`sin45+sin75=2sin60cos(-15)=2sin60cos15`

`"to obtain the exact value for "cos15`

`cos15=cos(45-30)=cos45cos30+sin45sin30`

`color(white)(cos15)=1/sqrt2xxsqrt3/2+1/sqrt2xx1/2`

`color(white)(cos15)=(sqrt3+1)/(2sqrt2)=1/4sqrt3(sqrt6+sqrt2)`

`rArrsin45+sin75=2xxsqrt3/2xx1/4(sqrt6+sqrt2)`

`color(white)(xxxxxxxxxxxx)=1/4sqrt3(sqrt6+sqrt2)`

`rArr(cos45-cos75)/(sin45+sin75)`

`=(cancel(1/4sqrt3)(sqrt6-sqrt2))/(cancel(1/4sqrt3)(sqrt6+sqrt2))`

`=(sqrt6-sqrt2)/(sqrt6+sqrt2)xx(sqrt6-sqrt2)/(sqrt6-sqrt2)larrcolor(blue)"rationalise denominator"`

`=(6-2sqrt12+2)/(6-2)=(8-2(2sqrt3))/4=2-sqrt3`