find the value 3(sec75)^2 with no calculator hint 75=45+30?

Question

This is my work were did i go wrong?
3/(cos45cos30-sin45sin30)^2 =
3/( (1/(sqrt2))*((sqrt3)/2)-(1/(sqrt2))*(1/2))^2=
3/(((sqrt3)-1) / (2(sqrt2)))^2 =
(3/1) * ((2(sqrt2))/((sqrt3)-1))^2 =
(3*4*2)/(3-2(sqrt3)+1) =
24/(4-2(sqrt3)) = 44.785
this is what i did and i got it wrong can you guys see where i went wrong ? thanks

Answer 1

3(sec(75))^2
3(1/cos(75))^2
3/(cos(75)^2)

cos(x + y) = cos(x)cos(y) - sin(x)sin(y)

cos(75) = cos(45 + 30)

cos(45 + 30) = cos(45)cos(30) - sin(45)sin(30)
cos(45 + 30) = ((sqrt(2)/2)(sqrt(3)/2)) - ((sqrt(2)/2)(1/2)))
cos(45 + 30) = (sqrt(6)/4) - (sqrt(2)/4)
cos(45 + 30) = (sqrt(6) - sqrt(2))/4

3/((sqrt(6) - sqrt(2))/4)^2
3/(((sqrt(6) - sqrt(2))(sqrt(6) - sqrt(2))/16)
(3/1)/((6 - sqrt(12) - sqrt(12) + 2)/16)
(3/1)*(16/(8 - 2sqrt(4 * 3)))
48/(8 - 4sqrt(3))
48/(4(2 - sqrt(3))
12/(2 - sqrt(3))

Multiply top and bottom by (2 + sqrt(3))

(12(2 + sqrt(3))/((2 - sqrt(3))(2 + sqrt(3))
(12(2 + sqrt(3))/(4 - 3)
(12(2 + sqrt(3))/1

ANS : 24 + 12sqrt(3)

if you plug in 3(1/cos(75))^2 and then divide it by 24 + 12sqrt(3), you will get 1, which means that they are the same value.

To take your answer

24/(4 - 2sqrt(3)) =
24/(2(2 - sqrt(3))) =
12/(2 - sqrt(3)) =
(12(2 + sqrt(3)))/((2 - sqrt(3))(2 + sqrt(3))
(12(2 + sqrt(3)))/(4 - 3)
(12(2 + sqrt(3)))/1

ANS : 24 + 12sqrt(3)

so all you had to do was simplify your answer.

Answer 2

you are correct if you mean 75 degrees

but you can simplify further
like

24/(4-2(sqrt3)) = 12/[2-sqrt3] {cancelling off 2}

=12*[2+sqrt3] / {[2+sqrt3][2-sqrt3]} introducing conjugate

=12*[2+sqrt3] /(4-3)

=12*[2+sqrt3]

= 24 + 12*sqrt3

= 24 +12*1.7321

=24 +20.7852

=44.7852 (no calculator needed)

Answer 3

cos45cos30-sin45sin30

= (sqrt(3)-1/2*sqrt(2))

It's square is (2-sqrt(2))/4

That will give you 12/(2-sqrt(2))

Simplify to get the correct answer.

Answer 4

I look at that and all I see is a jumble of numbers. Makes no sense to me whatsoever. What is the use of all that, anyway??

Answer 5

I dunno....... 44.7846 *is* the correct numerical value and it looks as if you got to it the correct way.

I'd say ask your Prof. Sometimes they have first year TA's who do their grading and make mistakes.


Doug

Answer 6

hey i love ur hint.