Trigonometry: Solved Examples

Finding all ratios from one

Given $\sec\theta = \dfrac{13}{12}$, find all other trigonometric ratios.

Solution: $\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}}$, so take hypotenuse = 13 and adjacent = 12. By the Pythagorean theorem, opposite = $\sqrt{13^2 - 12^2} = 5$. That gives:

An identity check

If $3\cot A = 4$, check whether $\dfrac{1-\tan^2 A}{1+\tan^2 A}$ equals $\cos^2 A - \sin^2 A$.

Solution: $\cot A = \dfrac{4}{3}$, so $\tan A = \dfrac{3}{4}$: a 3-4-5 triangle gives $\sin A = \dfrac{3}{5}$, $\cos A = \dfrac{4}{5}$.

Both sides equal $\dfrac{7}{25}$, so the identity holds here.

Evaluating a standard-angle expression

Evaluate $\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 60^\circ}$.

Solution: $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$, and since $\cos 30^\circ = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$, both $\sec 30^\circ$ and $\csc 60^\circ$ equal $\dfrac{2}{\sqrt{3}}$:

Solving for two unknown angles

If $\tan(A+B) = \sqrt{3}$ and $\tan(A-B) = \dfrac{1}{\sqrt{3}}$, with $0^\circ < A+B \le 90^\circ$ and $A > B$, find $A$ and $B$.

Solution: $\tan(A+B) = \tan 60^\circ \Rightarrow A+B = 60^\circ$, and $\tan(A-B) = \tan 30^\circ \Rightarrow A-B = 30^\circ$. Adding the two equations gives $2A = 90^\circ$, so $A = 45^\circ$ and $B = 15^\circ$.

Co-function identities

Every trig ratio has a complementary partner: $\sin(90^\circ-\theta)=\cos\theta$, $\tan(90^\circ-\theta)=\cot\theta$, and $\csc(90^\circ-\theta)=\sec\theta$. Rewriting one ratio in terms of its complement often makes an expression collapse.

Evaluate $\dfrac{\sin 18^\circ}{\cos 72^\circ}$.

Solution: Since $18^\circ = 90^\circ - 72^\circ$, $\sin 18^\circ = \cos 72^\circ$, so the ratio is $\dfrac{\cos 72^\circ}{\cos 72^\circ} = 1$.

Proving an identity

Prove that $\left(\csc\theta - \cot\theta\right)^2 = \dfrac{1-\cos\theta}{1+\cos\theta}$.

Solution: Write both terms on the left over a common denominator, then simplify using the difference of squares:

Worked example: a product-to-sum calculation

Calculate $\sin 65^\circ \sin 25^\circ$.

Solution: Since $25^\circ = 90^\circ - 65^\circ$, $\sin 25^\circ = \cos 65^\circ$, turning the product into $\sin 65^\circ \cos 65^\circ$. Using the double-angle identity $2\sin\theta\cos\theta = \sin 2\theta$: