Trigonometric Functions - Trigonometric Functions of Sum and Difference of Two Angles

Find the exact value of $\sin 15^\circ$.

Step 1: Express $15^\circ$ as a difference of two standard angles: $15^\circ = 45^\circ - 30^\circ$. \nStep 2: Use the formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$. \nStep 3: Substitute $A = 45^\circ$ and $B = 30^\circ$: $\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$ \nStep 4: Plug in the standard values: $\sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)$ \nStep 5: Simplify the expression: $\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

This approach decomposes a non-standard angle into standard angles whose trigonometric values are known from the unit circle, then applies the sine difference identity.

Prove that $\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ$.

Step 1: Take the Left Hand Side (LHS) and divide both numerator and denominator by $\cos 11^\circ$: $\text{LHS} = \frac{\frac{\cos 11^\circ}{\cos 11^\circ} + \frac{\sin 11^\circ}{\cos 11^\circ}}{\frac{\cos 11^\circ}{\cos 11^\circ} - \frac{\sin 11^\circ}{\cos 11^\circ}}$ \nStep 2: Simplify using $\tan \theta = \frac{\sin \theta}{\cos \theta}$: $\text{LHS} = \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ}$ \nStep 3: Recognize that $1 = \tan 45^\circ$. Substitute this into the expression: $\text{LHS} = \frac{\tan 45^\circ + \tan 11^\circ}{1 - (\tan 45^\circ)(\tan 11^\circ)}$ \nStep 4: Observe that this matches the form $\frac{\tan A + \tan B}{1 - \tan A \tan B} = \tan(A + B)$. \nStep 5: Therefore, $\text{LHS} = \tan(45^\circ + 11^\circ) = \tan 56^\circ$. \nStep 6: LHS = RHS. Proved.

This example uses the tangent sum formula in reverse. By dividing by $\cos A$, we transform a sine-cosine fraction into a tangent expression, which is a common technique in trigonometric proofs.

We can write $75^\circ$ as $45^\circ + 30^\circ$. Using the formula: $\cos(A + B) = \cos A \cos B - \sin A \sin B$ Substitute $A = 45^\circ$ and $B = 30^\circ$: $\cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ$ Substitute the standard values: $\cos 75^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)$ $\cos 75^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}$ To rationalize the denominator: $\cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$

This example demonstrates how to decompose a non-standard angle into the sum of two standard angles ($30^\circ$, $45^\circ$) to apply the cosine addition formula.

We can express $105^\circ$ as $60^\circ + 45^\circ$. Using the formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ Substitute $A = 60^\circ$ and $B = 45^\circ$: $\tan(60^\circ + 45^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ}$ Since $\tan 60^\circ = \sqrt{3}$ and $\tan 45^\circ = 1$: $\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}(1)}$ $\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$ Multiplying numerator and denominator by $(1 + \sqrt{3})$: $\tan 105^\circ = \frac{(\sqrt{3} + 1)^2}{1 - 3} = \frac{3 + 1 + 2\sqrt{3}}{-2}$ $\tan 105^\circ = \frac{4 + 2\sqrt{3}}{-2} = -(2 + \sqrt{3})$

By breaking $105^\circ$ into two special angles, we can use the tangent sum identity and rationalize the resulting radical expression.