Given sin23=p

Question

Determine the following in terms of p without using a calculator
1. Sin 67
2.tan 23
3.sin7
4.cos44

Answer 1

cos θ = ± √ 1- sin² θ

23° lies in the first square where all trigonometric functions are positive so:

cos 23° = √ [ 1 - sin² ( 23° ) ]

cos 23° = √ ( 1 - p² )

1.

67° = 90° - 23°

sin ( 90° - θ ) = cos θ

sin 67° = cos 23° = √ ( 1 - p² )

2.

tan 23° = sin 23° / cos 23° = p / √ ( 1 - p² )

3.

7° = 30° - 23°

sin ( A - B ) = sin A ∙ cosB - cos A ∙ sin B

sin 7° = sin ( 30° - 23° ) = sin 30° ∙ cos 23° - cos 30° ∙ sin 23°

sin 7° = 1 / 2 ∙ √ ( 1 - p² ) - √ 3 / 2 ∙ p = [ √ ( 1 - p² ) - √ 3 p ] / 2

4.

44° = 90° - 46°

cos ( 90° - θ ) = sin θ

cos 44° = cos ( 90° - 46° ) = sin 46°

46° = 2 ∙ 23°

sin ( 2θ ) = 2 ∙ sin θ ∙ cos θ

sin 46° = sin ( 2 ∙ 23° ) = 2 ∙ sin 23° ∙ cos 23° = 2 ∙ p ∙ √ ( 1- p² )

cos 44° = sin 46°

cos 44° = 2 ∙ p ∙ √ ( 1 - p² )

Answer 2

sin23=p = p/1, make a sketch of a right-angled triangle with opposite p and hypotenuse 1
x^2 + y^2 = 1
x^2 + p^2= 1
x = √(1 - p^2) ,
cos 23 = √(1-p^2)

sin 67° = cos 23° = √(1-p^2)

tan 23° = sin23/cos23 = p/√(1-p^2)

sin30 = sin(7+23)
= sin7cos23 + cos7sin23
= sin7(√(1-p^2)) + cos7(p) = 1/2
also: sin 60 = sin(67-7)
= sin67cos7 - cos67sin7 = √3/2
OK, I am stuck here!

trying this:
tan 30 = tan(7+23) = (tan7 + tan23) / (1 - tan7tan23)
let x = tan7
1/√3 = (x + tan23) / (1 - xtan23)
√3 x + √3tan23 = 1 - xtan23
x(√3 + tan23) = 1 - √3tan23
x = (1 - √3tan23)/(1 + √3tan23) , and we know tan23!!
but .... that only gives me tan7

argghhhhh!!

for the last one:
cos 44 = cos(67 - 23)
= cos67cos23 + sin67sin23
= sin23cos23 + cos23sin23
= 2sin23cos23
= 2p√(1-p^2)

I am 100% sure of the results except I could not finish sin7, but what I have so far is correct
I checked with the actual value of sin23 for all

Answer 3

Yeahhh ! Bosnian got it !!!
I like his method , I guess I couldn't see the forest for the trees .