Question

1.`cos^2(π/24)+cos^2((19π)/24)+cos^2((31π)/24)+cos^2((37π)/24)=`? solve this

Answer 1

`cos^2(π/24)+cos^2({19π}/24)+cos^2({31π}/24)+cos^2({37π}/24) = 2`

Explanation:

Fun. I don't know how to do this one offhand, so we'll just try some things.

There don't appear to be complementary or supplementary angles obviously in play, so perhaps our best move is to start with the double angle formula.

`cos 2 theta = 2 \cos ^2 theta - 1`

`cos^2 theta = 1/2 ( 1 + cos 2 theta)`

`cos^2(π/24)+cos^2({19π}/24)+cos^2({31π}/24)+cos^2({37π}/24)`

`= 4(1/2) + 1/2( cos (pi/12) + cos ({19 pi}/12) + cos({ 31\pi}/12) + cos ({37 pi}/12) )`

Now we replace angles with coterminal ones (ones with the same trig functions) by subtracting `2 pi.`

`= 2 + 1/2( cos (pi/12) + cos ({19 pi}/12 -2pi ) + cos({ 31\pi}/12 - 2pi) + cos ({37 pi}/12 - 2pi) )`

`= 2 + 1/2( cos (pi/12) + cos (-{5pi}/12) + cos({ 7pi}/12 ) + cos ({13 pi}/12) )`

Now we replace angles by supplementary angles, which negates the cosine. We drop the minus sign in the cosine argument too which doesn't change the cosine.

`= 2 + 1/2( cos (pi/12) + cos ({5pi}/12 ) - cos (pi - { 7pi}/12 ) - cos (pi - {13 pi}/12) )`

`= 2 + 1/2( cos (pi/12) + cos ({5pi}/12) - cos({ 5pi}/12 ) - cos(-pi/12) )`

`= 2 + 1/2( cos (pi/12) + cos ({5pi}/12) - cos({ 5pi}/12 ) - cos(pi/12) )`

`= 2 + 1/2(0)`

`= 2`

Answer 2

`2`

Explanation:

We know that,

`cos(pi/2+theta)=-sintheta=>color(red)(cos^2(pi/2+theta)=(-sintheta)^2=sin^2theta`

So,

`color(red)(cos^2((31pi)/24)=cos^2(pi/2+(19pi)/24)=sin^2((19pi)/2)...to(1)`

`and cos((3pi)/2+theta)=sintheta=>color(blue)(cos^2((3pi)/2+theta)=sin^2theta`

`=>color(blue)(cos^2((37pi)/2)=cos^2((3pi)/2+pi/24)=sin^2(pi/24)...to(2)`

Using `(1) and (2)`

`X=cos^2(π/24)+cos^2((19π)/24)+color(red)(cos^2((31π)/24))+color(blue)(cos^2((37π)/24)`

`=cos^2(pi/24)+cos^2((19pi)/2)+color(red)(sin^2((19pi)/2))+color(blue)(sin^2(pi/24)`

`={cos^2(pi/24)+sin^2(pi/24)}+{cos^2((19pi)/2)+sin^2((19pi)/2}`

`=1+1...to[as, sin^2theta+cos^2theta=1]`

`=2`