Trigonometry - Trigonometric Identities of Sum and Difference of Angles

Find the exact value of $\sin 75^\circ$ using sum and difference identities.

1. Express $75^\circ$ as a sum of two standard angles: $75^\circ = 45^\circ + 30^\circ$
2. Apply the sine sum identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $\sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$
3. Substitute the standard values: $\sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)$
4. Simplify the expression: $\sin 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

To solve for an angle not on the standard unit circle, we decompose it into the sum of two angles whose sine and cosine values are known ($45^\circ$ and $30^\circ$). We then apply the sine sum formula and simplify the fractions.

Prove that $\frac{\cos 8^\circ - \sin 8^\circ}{\cos 8^\circ + \sin 8^\circ} = \tan 37^\circ$.

1. Start with the Left Hand Side (LHS) and divide both numerator and denominator by $\cos 8^\circ$: $LHS = \frac{\frac{\cos 8^\circ}{\cos 8^\circ} - \frac{\sin 8^\circ}{\cos 8^\circ}}{\frac{\cos 8^\circ}{\cos 8^\circ} + \frac{\sin 8^\circ}{\cos 8^\circ}} = \frac{1 - \tan 8^\circ}{1 + \tan 8^\circ}$
2. Recognize that $1 = \tan 45^\circ$. Substitute this into the formula: $LHS = \frac{\tan 45^\circ - \tan 8^\circ}{1 + (\tan 45^\circ)(\tan 8^\circ)}$
3. This expression matches the structure of the identity $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$: $LHS = \tan(45^\circ - 8^\circ)$
4. Simplify the subtraction: $LHS = \tan 37^\circ = RHS$

This is a common algebraic manipulation in trigonometry. By dividing by $\cos \theta$, we convert a sine-cosine expression into a tangent expression, which allows us to use the $\tan(A-B)$ identity by recognizing that $1$ is $\tan 45^\circ$.

We can write $105^{\circ}$ as $60^{\circ} + 45^{\circ}$. Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$: $\cos(60^{\circ} + 45^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ}$ Substitute the standard values: $\cos 105^{\circ} = \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right) - \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}}\right)$ $\cos 105^{\circ} = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}}$ $\cos 105^{\circ} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$

To find the value of a non-standard angle, we decompose it into the sum of two standard angles (60 and 45) and apply the cosine sum identity.

We can write $15^{\circ}$ as $45^{\circ} - 30^{\circ}$. Using the identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$: $\tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}}$ Substitute $\tan 45^{\circ} = 1$ and $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$: $\tan 15^{\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ Rationalize the denominator: $\tan 15^{\circ} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1}$ $\tan 15^{\circ} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$ Hence proved.

By representing 15 degrees as a difference of 45 and 30, we use the tangent difference formula and simplify the resulting radical expression through rationalization.