Question

Use an Addition orSubtraction Formula to find the exact value of the expression, asDemonstrated$\mathbf{tan}\mathbf{}{\mathbf{15}}^{\mathbf{\circ }}$

Short answer

Using subtraction formula of sine. We get$\mathbf{tan}{\mathbf{15}}^{\mathbf{\circ }}\mathbf{=}{\mathbf{(}\sqrt{\mathbf{3}}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}$

Step-by-step solution

Step 1. Given information.

We know that sine and cosine are the trigonometric functions, Where x and y

are the angles

subtraction formula of a tangent, we can easily calculate the value of $\tan {15}^{\circ }$.

Formula$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

Step 2. Determining the value.

Split the angle of $\tan {15}^{\circ }$,

That is ${15}^{\circ }={45}^{\circ }-{30}^{\circ }$, Since we know the exact values of sine and cosine at 45 and 30

So, we got the values of x and y, $x={45}^{\circ }$

And $y={30}^{\circ }$

Now apply the addition formula of sine to get localid="1650448502233" $\tan {15}^{\circ }$

By using $\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

$\tan {15}^{\circ }=\tan ({45}^{\circ }-\tan {30}^{\circ })$

$\tan \left({15}^{\circ }\right)=\frac{\tan {45}^{\circ }-\tan {30}^{\circ }}{1+\tan {45}^{\circ }\times \tan {30}^{\circ }}$

$\tan \left({15}^{\circ }\right)=\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}$

$\tan \left({15}^{\circ }\right)=\frac{(\sqrt{3}-1)}{\sqrt{3}+1}$

localid="1651059191978" $\tan \left({15}^{\circ }\right)={(\sqrt{3}-1)}^2$