Given sin23=p
Determine the following in terms of p without using a calculator
1. Sin 67
2.tan 23
3.sin7
4.cos44
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 )
Yeahhh ! Bosnian got it !!!
I like his method , I guess I couldn't see the forest for the trees .
To determine the values without using a calculator, let's break down each question and use trigonometric identities and formulas:
1. Sin 67:
Since sin 23 = p, we can find sin 67 using the double-angle identity for sine. The double-angle identity states that:
sin(2θ) = 2sin(θ)cos(θ)
Let's use this formula to find sin 67:
sin(67) = sin(2 * 33.5) = 2 sin(33.5) cos(33.5)
We need to find sin 33.5 and cos 33.5. Since we know that sin 23 = p:
sin(33.5) = sin(23 + 10.5) = sin23 * cos10.5 + cos23 * sin10.5 = p * cos(10.5) + cos(23) * sin(10.5)
Now we need to find cos 10.5. To do this, we can use the half-angle identity for cosine:
cos(2θ) = cos^2(θ) - sin^2(θ)
cos(10.5) = √(1 - sin^2(10.5))
We can find sin(10.5) using the previous approach:
sin(10.5) = sin(23 - 12.5) = sin(23) * cos(12.5) - cos(23) * sin(12.5)
After finding sin(10.5), we can substitute it into the formula for cos(10.5) to get the final value.
Finally, substitute all the values back into sin(67):
sin(67) = 2 * (p * cos(10.5) + cos(23) * sin(10.5)) * cos(33.5)
2. Tan 23:
Similar to finding sin 67, we can use the identity for tangent:
tan(2θ) = (2tan(θ))/(1 - tan^2(θ))
tan(23) = tan(2 * 11.5) = (2tan(11.5))/(1 - tan^2(11.5))
Since we know that sin 23 = p, we can use this to find tan 11.5:
tan(11.5) = sin(11.5)/cos(11.5) = (1 - cos^2(11.5))/cos(11.5)
Now, substitute this into the formula for tan(23) to get the final value.
3. Sin 7:
Similarly, we can use the double-angle identity for sine to find sin 7:
sin(7) = sin(2 * 3.5) = 2 sin(3.5) cos(3.5)
To find sin(3.5), we can use the half-angle identity for sine.
4. Cos 44:
We can use the complementary angle identity for sine to find cos 44:
cos(44) = sin(90 - 44) = sin(46)
To find sin(46), we can follow the same steps as in question 3 to find sin 46 using the half-angle identity.
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