Problem 33
Question
In Exercises \(29-36,\) write the expression as the sine, cosine, or tangent of an angle. $$ \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}} $$
Step-by-Step Solution
Verified Answer
The given expression simplifies to \( \tan 15^\circ \)
1Step 1: Recognize the Given Pattern
Realize that the given expression \(\dfrac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}\) fits into the pattern of an identity for the tangent of the difference of two angles, \(\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}\).
2Step 2: Identify the angles
From the pattern it is realized that angle A is \(45^\circ\) and angle B is \(30^\circ\). So, the given expression simplifies to \(\tan(A - B)\) which means \(\tan(45^\circ - 30^\circ)\)
3Step 3: Simplify the expression
Subtract the angles to simplify the expression. The given expression is then equivalent to \(\tan 15^\circ\)
Other exercises in this chapter
Problem 32
In Exercises 9-50, verify the identity \( \cos x - \dfrac{\cos x}{1 - \tan x} = \dfrac{\sin x \cos x}{\sin x - \cos x} \)
View solution Problem 33
In Exercises 29-36, use a double-angle formula to rewrite the expression. \( 4 - 8 \sin^2 x \)
View solution Problem 33
In Exercises 25-38, find all solutions of the equation in the interval \( [0, 2\pi) \). \( 2 \cos^2 x + \cos x - 1 = 0 \)
View solution Problem 33
In Exercises 9-50, verify the identity \( \tan \left(\dfrac{\pi}{2} - \theta \right) \tan \theta = 1 \)
View solution