Question

Use an Addition orSubtraction Formula to find the exact value of the expression, asDemonstrated$\mathbf{tan}\mathbf{}{\mathbf{165}}^{\mathbf{\circ }}$

Short answer

Using subtraction formula of sine. We get$\mathbf{tan}{\mathbf{165}}^{\mathbf{\circ }}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{\mathbf{3}}$

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 $\tan {165}^{\circ }$,

That is ${165}^{\circ }={45}^{\circ }+{120}^{\circ }$, 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 $y={120}^{\circ }$

Now apply the addition formula of sine to get$\tan {165}^{\circ }$

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

$\tan {165}^{\circ }=\tan ({45}^{\circ }+{120}^{\circ })$

$\tan \left({165}^{\circ }\right)=\frac{\tan {45}^{\circ }+\tan {120}^{\circ }}{1-\tan {45}^{\circ }\times \tan {120}^{\circ }}$

$\tan \left({165}^{\circ }\right)=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$\tan \left({165}^{\circ }\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}=-2+\sqrt{3}$