Problem 13 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the exact values of the sine, cosine, and tangent of the angle. $$105^{\circ}=60^{\circ}+45^{\circ}$$

Step-by-Step Solution

Answer

The exact values of the sine, cosine, and tangent of \(105^{\circ}\) are \(\frac{\sqrt{6} + \sqrt{2}}{4}\), \(\frac{\sqrt{2} - \sqrt{6}}{4}\), and \(-\sqrt{2} + \sqrt{6}\), respectively.

Step 1: Recall the sum-of-angles formula

Recall the following formulas: \[\sin (A+B) = \sin A \cos B + \cos A \sin B\], \[\cos (A+B) = \cos A \cos B - \sin A \sin B\], and \[\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]. You'll use A as \(60^{\circ}\) and B as \(45^{\circ}\)

Step 2: Calculate the sine

Substitute A and B into the formula for sin(A+B): \[\sin (105^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ}\]. This equals \[\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} + \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\]

Step 3: Calculate the cosine

Substitute A and B into the formula for cos(A+B): \[\cos (105^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ}\]. This equals \[\frac{1}{2} \times \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4}\]

Step 4: Calculate the tangent

The formula for tan(A+B) becomes a little tricky, as it involves division. But because tan is just the ratio of sin to cos, you can simply divide the results from Step 2 and Step 3: \[\tan (105^{\circ}) = \frac{\sin 105^{\circ}}{\cos 105^{\circ}} = \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{2} - \sqrt{6}}{4}}\] The 4's cancel, and you're left with the ratio \(-\sqrt{2} + \sqrt{6}\)