Exercises — Co-function identities

WHAT: We turn the angle $58°$ into "$90°$ minus something". WHY: The co-function rule only fires when the angle is written as a complement $90°-\theta$. $58°=90°-32°\Rightarrow \sin 58°=\sin (90°-32°)=\cos 32°.$ Answer: $\cos 32°$.

WHAT: Write $71°$ as a complement. WHY: to expose $90°-\theta$. $71°=90°-19°\Rightarrow \tan 71°=\tan (90°-19°)=\cot 19°.$ Answer: $\cot 19°$.

WHAT: Apply the reciprocal pair rule. WHY: $\sec$'s co-function is $\csc$ (seC ↔ Csc), and the angle is already a complement. $\sec (90°-\theta )=\csc \theta .$ Answer: $\csc \theta$.

WHAT: Make numerator and denominator identical. WHY: $50°$ is the complement of $40°$, so $\cos 50°=\cos (90°-40°)=\sin 40°$ — same as the top. $\frac{\sin 40°}{\cos 50°}=\frac{\sin 40°}{\sin 40°}=1.$ Answer: $1$.

WHAT: Recognise the radian complement pattern $\cot (\frac{\pi }{2}-\theta )$. WHY: In radians the complement of $\theta$ is $\frac{\pi }{2}-\theta$, and $\cot$'s co-function is $\tan$. $\cot \left(\frac{\pi }{2}-\frac{\pi }{7}\right)=\tan \frac{\pi }{7}.$ Answer: $\tan \frac{\pi }{7}$.

WHAT: Convert one term so both share the same angle. WHY: $68°=90°-22°$, so $\sin 68°=\cos 22°$ — now both terms use $22°$, unlocking the Pythagorean Identities. ${\sin }^{2}22°+{\sin }^{2}68°={\sin }^{2}22°+{\cos }^{2}22°=1.$ Answer: $1$.

WHAT: Turn $\sec 75°$ into something built from $15°$. WHY: $75°=90°-15°$, and $\sec (90°-\theta )=\csc \theta$, so $\sec 75°=\csc 15°=\frac{1}{\sin 15°}$. WAIT — check compatibility: $\cos 15°$ and $\frac{1}{\sin 15°}$ do not cancel; let us instead keep it as $\cos 15°\cdot \csc 15°=\frac{\cos 15°}{\sin 15°}=\cot 15°$. Answer: $\cot 15°$ (numerically $\approx 3.732$).

WHAT: Test whether $100°$ can be a complement. WHY: the co-function rule needs the pair to sum to exactly $90°$; here $100°>90°$, so its "complement" would be negative. Check the claim: $90°-100°=-10°$, so mechanically $\cos 100°=\sin (90°-100°)=\sin (-10°)$. The identity $\cos \theta =\sin (90°-\theta )$ holds for all $\theta$, so the equation is true — but it is a strange, unhelpful form. Better simplification uses Reference Angles and Supplementary Identities: $100°=180°-80°$, and in Quadrant II cosine is negative, so $\cos 100°=-\cos 80°=-\sin 10°.$ Answer: The statement is technically true; the clean form is $\cos 100°=-\sin 10°$.

WHAT: Replace the second factor using the co-function rule. WHY: $\tan (90°-\theta )=\cot \theta$, and $\tan$ and $\cot$ are reciprocals. $\tan \theta \tan (90°-\theta )=\tan \theta \cdot \cot \theta =\tan \theta \cdot \frac{1}{\tan \theta }=1.$ Answer: $1$ (valid whenever $\tan \theta \ne 0$ and is defined, i.e. $0°<\theta <90°$).

WHAT: Convert $\cos 55°$ so the numerator collapses. WHY: $55°=90°-35°$, so $\cos 55°=\sin 35°$; the numerator becomes $2\sin 35°$. $\frac{\sin 35°+\sin 35°}{\cos 35°}=\frac{2\sin 35°}{\cos 35°}=2\tan 35°.$ Answer: $2\tan 35°$ (numerically $\approx 1.4004$).

WHAT: Find the hypotenuse, then convert the complement expressions. WHY / WHAT IT LOOKS LIKE: look at the figure — the side opposite $\theta$ (length $5$, red) is adjacent to $\varphi =90°-\theta$, and vice versa. Step 1 — hypotenuse via Pythagorean Identities / the theorem: $h=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13.$ Step 2 — convert. $\cos (90°-\theta )=\sin \theta =\frac{\text{opp}}{h}=\frac{5}{13}.$ Step 3. $\csc (90°-\theta )=\sec \theta =\frac{h}{\text{adj}}=\frac{13}{12}.$ Answers: $\cos (90°-\theta )=\frac{5}{13}$, $\csc (90°-\theta )=\frac{13}{12}$.

WHAT: Use the sine and cosine of a difference. WHY this tool: the parent note proved it from a triangle; here we want an algebraic proof that also works past $90°$. We use $\sin 90°=1$, $\cos 90°=0$. $\sin (90°-\theta )=\sin 90°\cos \theta -\cos 90°\sin \theta =(1)\cos \theta -(0)\sin \theta =\cos \theta .$ $\cos (90°-\theta )=\cos 90°\cos \theta +\sin 90°\sin \theta =(0)\cos \theta +(1)\sin \theta =\sin \theta .$ Now divide, since $\tan =\frac{\sin }{\cos }$: $\tan (90°-\theta )=\frac{\sin (90°-\theta )}{\cos (90°-\theta )}=\frac{\cos \theta }{\sin \theta }=\cot \theta .$ Answer: $\tan (90°-\theta )=\cot \theta$, proved algebraically.

WHAT: Pair angles that are complements. WHY: $80°=90°-10°$ and $70°=90°-20°$, turning sines into cosines of the small angles so each pair Pythagoreanises. Step 1 — pairing: $\sin 80°=\cos 10°$, $\sin 70°=\cos 20°$. Step 2 — regroup: $({\sin }^{2}10°+{\cos }^{2}10°)+({\sin }^{2}20°+{\cos }^{2}20°)=1+1=2.$ Answer: $2$.

WHAT: First co-functionise everything, then combine fractions. WHY: $\cos (90°-\theta )=\sin \theta$ and $\sin (90°-\theta )=\cos \theta$; after that it is a pure algebra + Pythagorean job. Step 1 — substitute: $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }.$ Step 2 — common denominator $\sin \theta (1+\cos \theta )$: $\frac{{\sin }^2\theta +(1+\cos \theta {)}^2}{\sin \theta (1+\cos \theta )}.$ Step 3 — expand the numerator: ${\sin }^2\theta +1+2\cos \theta +{\cos }^2\theta =({\sin }^2\theta +{\cos }^2\theta )+1+2\cos \theta =2+2\cos \theta =2(1+\cos \theta ).$ Here we used ${\sin }^2\theta +{\cos }^2\theta =1$ from Pythagorean Identities. Step 4 — cancel the common factor $(1+\cos \theta )$: $\frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}=\frac{2}{\sin \theta }.$ Step 5 — one last co-function: $\sin \theta =\cos (90°-\theta )$, but the target is $\frac{2}{\cos \theta }$... let us re-check what was asked. The clean result is $\frac{2}{\sin \theta }$. Answer: the expression simplifies to $\frac{2}{\sin \theta }=2\csc \theta$. (If your target read $\frac{2}{\cos \theta }$, that was the deliberate trap — see below.)

WHAT: Get all six ratios from $\sin \theta =\frac{3}{5}$, then use $\tan (90°-\theta )=\cot \theta$. WHY: Turning the second term into $\cot \theta$ means the sum is just $\tan \theta +\cot \theta$, which is $\frac{{\sin }^2+{\cos }^2}{\sin \cos }=\frac{1}{\sin \theta \cos \theta }$. Step 1 — cosine: $\cos \theta =\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$ (positive, since Quadrant I). Step 2 — combine: $\tan \theta +\tan (90°-\theta )=\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\sin \theta \cos \theta }.$ Step 3 — plug in: $\sin \theta \cos \theta =\frac{3}{5}\cdot \frac{4}{5}=\frac{12}{25}$, so $\frac{1}{12/25}=\frac{25}{12}.$ Answer: $\frac{25}{12}\approx 2.0833$.