Question

100 Points! Algebra question. Photo attached. Find the exact value of the expression. Please show as much work as possible. Thank you!

Answer 1

`\tan 15^(\circ) = 2 - √(3)`

Explanation:

To find the exact value of tan 15°, we can use trigonometric identities and the unit circle.

We know that tan(x) can be expressed as the ratio of sin(x) and cos(x). We can also write 15° as (60° - 45°).

Therefore, tan 15° can be expressed as:

`\tan15^(\circ)=\tan(60^(\circ)-45^(\circ))=(\sin(60^(\circ)-45^(\circ)))/(\cos(60^(\circ)-45^(\circ)))`

Now use the trigonometric angle identities to rewrite the ratio in terms of sin 60°, cos 60°, sin 45° and cos 45°.

`\boxed{\begin{minipage}{6.5 cm}\underline{Trigonometric Angle Identities}\\\\$\sin (A - B)=\sin A \cos B - \cos A \sin B$\\\\$\cos (A - B)=\cos A \cos B + \sin A \sin B$\\\end{minipage}}`

Therefore:

`\begin{aligned}\tan15^(\circ)&=(\sin(60^(\circ)-45^(\circ)))/(\cos(60^(\circ)-45^(\circ)))\\\\&=(\sin60^(\circ)\cos45^(\circ)-\cos60^(\circ)\sin45^(\circ))/(\cos 60^(\circ) \cos 45^(\circ)+ \sin 60^(\circ)\sin 45^(\circ))\end{aligned}`

In the unit circle, the cosine of an angle is represented by the x-coordinate of a point on the circle, and the sine of an angle is represented by the y-coordinate of that same point → (x, y) = (cos θ, sin θ). Therefore, we can use the unit circle to identity the values of sin 60°, cos 60°, sin 45° and cos 45°:

`\sin 60^(\circ)=(√(3))/(2)`

`\cos 60^(\circ)=(1)/(2)`

`\sin 45^(\circ)=(√(2))/(2)`

`\cos 45^(\circ)=(√(2))/(2)`

Substitute these into the equation and simplify:

`\begin{aligned}\tan15^(\circ)&=(\sin(60^(\circ)-45^(\circ)))/(\cos(60^(\circ)-45^(\circ)))\\\\&=(\sin60^(\circ)\cos45^(\circ)-\cos60^(\circ)\sin45^(\circ))/(\cos 60^(\circ) \cos 45^(\circ)+ \sin 60^(\circ)\sin 45^(\circ))\\\\&=((√(3))/(2)\cdot (√(2))/(2)-(1)/(2)\cdot (√(2))/(2))/((1)/(2)\cdot (√(2))/(2)+ (√(3))/(2)\cdot (√(2))/(2))\\\\\end{aligned}`

`\begin{aligned}&=((√(2))/(2) \left((√(3))/(2)-(1)/(2)\right))/((√(2))/(2) \left((1)/(2)+ (√(3))/(2)\right))\\\\&=((√(3))/(2)-(1)/(2))/( (1)/(2)+ (√(3))/(2))\\\\&=((√(3)-1)/(2))/((1+√(3))/(2))\\\\&=(√(3)-1)/(1+√(3))\end{aligned}`

Simplify further by multiplying the numerator and denominator by the conjugate of the denominator:

`\begin{aligned}&=(√(3)-1)/(1+√(3))\cdot (1-√(3))/(1-√(3))\\\\&=((√(3)-1)(1-√(3)))/((1+√(3))(1-√(3)))\\\\&=(√(3)-3-1+√(3))/(1-√(3)+√(3)-3)\\\\&=(2√(3)-4)/(-2)\\\\&=-√(3)+2\\\\&=2-√(3)\end{aligned}`

Therefore, the exact value of tan 15° is (2 - √3).

Answer 2

`2-√(3)`

Explanation:

Find the exact value of the expression, tan(15°).

The method I am about to show you will allow you to solve this problem without any tables or calculators. Although, memorizing the unit circle and trigonometric identities is required.

`\tan(15 \textdegree)\\\\\Longrightarrow \tan((30 \textdegree)/(2) )\\\\\text{Use the half-angle identity:} \ \tan((A)/(2))=\pm \sqrt{(1-\cos(A))/(1+\cos(A)) }=(\sin(A))/(1+\cos(A)) =(1-\cos(A))/(\sin(A)) \\\\\Longrightarrow(1-\cos(30 \textdegree))/(\sin(30 \textdegree)) \\\\\text{From the unit circle:} \ \cos(30 \textdegree)=(√(3) )/(2) \ \text{and} \ \sin(30 \textdegree)=(1)/(2)\\`

`\Longrightarrow (1-(√(3) )/(2))/((1)/(2))\\\\\Longrightarrow 2(1-(√(3) )/(2))\\\\\therefore \boxed{\boxed{\tan(15 \textdegree)=2-√(3) }}`

Thus, the problem is solved.