Answer :
Answer:
[tex]\textsf{1)}\quad \sin(18^{\circ}) = 2k\sqrt{1 - k^2}[/tex]
[tex]\textsf{2)}\quad \cos\left(5^{\circ}\right)=\sqrt{\dfrac{1+p}{2}}[/tex]
[tex]\textsf{3)}\quad \sin\left(5^{\circ}\right)=\sqrt{\dfrac{1-p}{2}}[/tex]
[tex]\textsf{4)}\quad \cos (5^{\circ})=\dfrac{p+\sqrt{3-3p^2}}{2}[/tex]
Step-by-step explanation:
Question 1
To determine the value of sin 18° in terms of k, given that sin 9° = k, we can use the double-angle identity for sine:
[tex]\boxed{\begin{array}{c}\underline{\textsf{Sine double-angle identity}}\\\\\sin(2\theta) = 2\sin(\theta)\cos(\theta)\end{array}}[/tex]
Let θ = 9°. Therefore:
[tex]\sin(18^{\circ})=2\sin(9^{\circ})\cos (9^{\circ})[/tex]
To find cos 9°, use the Pythagorean identity sin²θ + cos²θ = 1:
[tex]\sin^2(9^{\circ}) + \cos^2(9^{\circ}) = 1\\\\\cos^2(9^{\circ})=1-\sin^2(9^{\circ})\\\\\cos(9^{\circ})=\pm\sqrt{1-\sin^2(9^{\circ})}\\\\\cos(9^{\circ})=\pm\sqrt{1-k^2}[/tex]
As 9° is in the first quadrant, where both the sine and cosine of the angle are positive, take the positive square root:
[tex]\cos(9^{\circ})=\sqrt{1-k^2}[/tex]
Finally, substitute the values of sin 9° and cos 9° in terms of k:
[tex]\Large\boxed{\boxed{\sin(18^{\circ}) = 2k\sqrt{1 - k^2}}}[/tex]
[tex]\dotfill[/tex]
Question 2
To determine the value of cos 5° in terms of p, given that cos 10° = p, we can use the half-angle formula for cosine:
[tex]\boxed{\begin{array}{c}\underline{\textsf{Cosine half-angle formula}}\\\\\cos\left(\dfrac{\theta}{2}\right)=\pm \sqrt{\dfrac{1+\cos \theta}{2}}\end{array}}[/tex]
Let θ = 10°. Therefore:
[tex]\cos\left(\dfrac{10^{\circ}}{2}\right)=\pm \sqrt{\dfrac{1+\cos 10^{\circ}}{2}}\\\\\\\cos\left(5^{\circ}\right)=\pm \sqrt{\dfrac{1+\cos 10^{\circ}}{2}}[/tex]
As 10° is in the first quadrant, where both the sine and cosine of the angle are positive, take the positive square root:
[tex]\cos\left(5^{\circ}\right)=\sqrt{\dfrac{1+\cos 10^{\circ}}{2}}[/tex]
Finally, substitute in cos 10° = p:
[tex]\Large\boxed{\boxed{\cos\left(5^{\circ}\right)=\sqrt{\dfrac{1+p}{2}}}}[/tex]
[tex]\dotfill[/tex]
Question 3
To determine the value of cos 5° in terms of p, given that cos 10° = p, we can use the half-angle formula for sine:
[tex]\boxed{\begin{array}{c}\underline{\textsf{Sine half-angle formula}}\\\\\sin\left(\dfrac{\theta}{2}\right)=\pm \sqrt{\dfrac{1-\cos \theta}{2}}\end{array}}[/tex]
Let θ = 10°. Therefore:
[tex]\sin\left(\dfrac{10^{\circ}}{2}\right)=\pm \sqrt{\dfrac{1-\cos 10^{\circ}}{2}}\\\\\\\sin\left(5^{\circ}\right)=\pm \sqrt{\dfrac{1-\cos 10^{\circ}}{2}}[/tex]
As 10° is in the first quadrant, where both the sine and cosine of the angle are positive, take the positive square root:
[tex]\sin\left(5^{\circ}\right)=\sqrt{\dfrac{1-\cos 10^{\circ}}{2}}[/tex]
Finally, substitute in cos 10° = p:
[tex]\Large\boxed{\boxed{\sin\left(5^{\circ}\right)=\sqrt{\dfrac{1-p}{2}}}}[/tex]
[tex]\dotfill[/tex]
Question 4
To determine the value of cos 5° in terms of p, given that cos 55° = p, begin by expressions sin 55° in terms of cos 55° using the Pythagorean identity sin²θ + cos²θ = 1:
[tex]\sin^2(55^{\circ})+\cos^2(55^{\circ})=1\\\\\\\sin^2(55^{\circ})=1-\cos^2(55^{\circ})\\\\\\\sin(55^{\circ})=\pm\sqrt{1-\cos^2(55^{\circ})}[/tex]
As 55° is in the first quadrant, where both the sine and cosine of the angle are positive, take the positive square root:
[tex]\sin(55^{\circ})=\sqrt{1-\cos^2(55^{\circ})}[/tex]
Given that cos 55° = p, then sin 55° expressed in terms of p is:
[tex]\sin(55^{\circ})=\sqrt{1-p^2}[/tex]
Now, find an expression for cos 5° by expressing 5° as (60° - 55°) and using the cosine difference formula:
[tex]\boxed{\begin{array}{c}\underline{\textsf{Cosine difference formula}}\\\\\cos(A-B)=\cos A \cos B +\sin A \sin B\end{array}}[/tex]
Therefore:
[tex]\cos (5^{\circ})=\cos (60^{\circ})\cos (55^{\circ})+\sin (60^{\circ})\sin (55^{\circ})[/tex]
As cos 60° = 1/2 and sin 60° = √3/2 then:
[tex]\cos (5^{\circ})=\dfrac{1}{2}\cos (55^{\circ})+\dfrac{\sqrt{3}}{2}\sin (55^{\circ})[/tex]
Substitute in the expressions for cos 55° and sin 55° in terms of p:
[tex]\cos (5^{\circ})=\dfrac{1}{2}p+\dfrac{\sqrt{3}}{2}\sqrt{1-p^2}[/tex]
Simplify:
[tex]\cos (5^{\circ})=\dfrac{p+\sqrt{3}\sqrt{1-p^2}}{2}[/tex]
[tex]\cos (5^{\circ})=\dfrac{p+\sqrt{3(1-p^2)}}{2}[/tex]
[tex]\Large\boxed{\boxed{\cos (5^{\circ})=\dfrac{p+\sqrt{3-3p^2}}{2}}}[/tex]
1) If sin9° = k, determine the value of sin 18° in terms of k:
Using the double angle identity for sine, we have:
sin(18°) = 2 * sin(9°) * cos(9°)
We know sin(9°) = k, and since cosine is the same as sine for complementary angles, we have:
cos(9°) = sqrt(1 - sin^2(9°)) = sqrt(1 - k^2)
So, sin(18°) = 2 * k * sqrt(1 - k^2).
2) If cos 10°= p, determine the value of cos 5° in terms of p:
Using the addition angle identity for cosine, we have:
cos(15°) = cos(10° + 5°) = cos(10°) * cos(5°) - sin(10°) * sin(5°)
We know cos(10°) = p, and sin(10°) = sqrt(1 - cos^2(10°)) = sqrt(1 - p^2), so:
cos(15°) = p * cos(5°) - sqrt(1 - p^2) * sin(5°)
To find cos(5°), let's solve this equation.
3) If cos 10° = p, determine the value of sin 5° in terms of p:
Using the Pythagorean identity for sine, we have:
sin(10°) = sqrt(1 - cos^2(10°))
We know cos(10°) = p, so:
sin(10°) = sqrt(1 - p^2)
Now, we can use the half-angle identity for sine:
sin(5°) = sqrt((1 - cos(10°)) / 2)
Substitute sin(10°) = sqrt(1 - p^2) into the equation:
sin(5°) = sqrt((1 - p^2) / 2)
4) If cos 55° = p, determine the value of cos 5° in terms of p:
Using the subtraction angle identity for cosine, we have:
cos(50°) = cos(55° - 5°) = cos(55°) * cos(5°) + sin(55°) * sin(5°)
We know cos(55°) = p, and sin(55°) = sqrt(1 - cos^2(55°)) = sqrt(1 - p^2), so:
cos(50°) = p * cos(5°) + sqrt(1 - p^2) * sin(5°)
To find cos(5°), let's solve this equation.
